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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Ncert solutions for class 10 maths chapter 8 – download free pdf.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry is helpful for the students as it aids in understanding the concepts as well as in scoring well in CBSE Class 10 board examination. The NCERT Solutions are designed and reviewed by subject experts and cover all the questions from the textbook. These NCERT Solutions are framed as per the latest update on the CBSE Syllabus for 2023-24 and its guidelines, in accordance with the exam pattern.

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The  NCERT Solutions for Class 10 Maths provide a strong foundation for every concept across all chapters. Students can clarify their doubts and understand the fundamentals of this chapter. Also, students can solve the difficult problems in each exercise with the help of these NCERT Solutions for Class 10  Maths Chapter 8.

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair of Linear Equations in Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction to Trigonometry
  • Chapter 9 Some Applications of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related to Circles
  • Chapter 13 Surface Areas and Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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NCERT Solutions for Class 10 Maths March28 Chapter 8 Introduction to Trigonometry

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Access Answers to NCERT Class 10 Maths Chapter 8 – Introduction to Trigonometry

Exercise 8.1 Page: 181

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC 2 =AB 2 +BC 2

AC 2  = (24) 2 +7 2

AC 2  = (576+49)

AC 2  = 625cm 2

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

trigonometry questions ncert

(ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

2. In Fig. 8.13, find tan P – cot R

Ncert solutions class 10 chapter 8-1

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

PR 2 = QR 2 + PQ 2

Substitute the values of PR and PQ

13 2 = QR 2 +12 2

169 = QR 2 +144

Therefore, QR 2 = 169−144

QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) = Opposite side /Adjacent side = QR/PQ = 5/12

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0

3. If sin A = 3/4, calculate cos A and tan A.

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC 2 =AB 2 + BC 2

Substitute the value of AC and BC

(4k) 2 =AB 2 + (3k) 2

16k 2 −9k 2 =AB 2

Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

Therefore, tan A = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Given: 15 cot A = 8

So, Cot A = 8/15

We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

Substitute the value of AB and BC

AC 2 = (8k) 2 + (15k) 2

AC 2 = 64k 2 + 225k 2

AC 2 = 289k 2

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

Sin (A) = Opposite side /Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get

Sin A = BC/AC = 15k/17k = 15/17

Therefore, sin A = 15/17

Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = Hypotenuse/Adjacent side

AC/AB = 17k/8k = 17/8

Therefore sec (A) = 17/8

trigonometry questions ncert

5. Given sec θ = 13/12 Calculate all other trigonometric ratios

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Substitute the value of AB and AC

(13k) 2 = (12k) 2 + BC 2

169k 2 = 144k 2 + BC 2

BC 2 = 169k 2 – 144k 2

BC 2 = 25k 2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD 2 = BC 2 – BD 2 … (3)

CD 2 =AC 2 −AD 2  ….(4)

From the equations (3) and (4) we get,

AC 2 −AD 2 = BC 2 −BD 2

Now substitute the equations (1) and (2) in (3) and (4)

K 2 (BC 2 −BD 2 )=(BC 2 −BD 2 ) k 2 =1

Putting this value in equation, we obtain

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

7. If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)

(ii) cot 2 θ

Let us assume a △ABC in which ∠B = 90° and ∠C = θ

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in △ABC we get.

AC 2 = AB 2 +BC 2

AC 2 = (8k) 2 +(7k) 2

AC 2 = 64k 2 +49k 2

AC 2 = 113k 2

AC = √113 k

According to the sine and cos function ratios, it is written as

sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113

Now apply the values of sin function and cos function:

Ncert solutions class 10 chapter 8-2

8. If 3 cot A = 4, check whether (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A or not.

Let △ABC in which ∠B=90°

We know that, cot function is the reciprocal of tan function and it is written as

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC 2 =(4k) 2 +(3k) 2

AC 2 =16k 2 +9k 2

AC 2 =25k 2

Now, apply the values corresponding to the ratios

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

Now compare the left hand side(LHS) with right hand side(RHS)

Ncert solutions class 10 chapter 8-3

Since, both the LHS and RHS = 7/25

R.H.S. =L.H.S.

Hence, (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A   is proved

9. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Let ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in ΔABC we get:

AC 2 =(√3 k) 2 +(k) 2

AC 2 =3k 2 +k 2

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Then find the values of cos C and sin C

Sin C = AB/AC = √ 3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = ( √ 3/2 )(1/2) – (1/2) ( √ 3/2 ) = 0

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

In a given triangle PQR, right angled at Q, the following measures are

PR + QR = 25 cm

Now let us assume, QR = x

According to the Pythagorean Theorem,

PR 2 = PQ 2 + QR 2

Substitute the value of PR as x

(25- x) 2 = 5 2 + x 2

25 2 + x 2 – 50x = 25 + x 2

625 + x 2 -50x -25 – x 2 = 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

Answer: False

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than 1.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC 2 =MN 2 +NC 2

5 2 =3 2 +4 2

25   =   25

(ii) sec A = 12/5 for some value of angle A

Answer: True

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

(12k) 2 =(5k) 2 +NC 2

NC 2 +25k 2 =144k 2

NC 2 =119k 2

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

Answer : False

Justification: sin θ = Opposite/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

Exercise 8.2 Page: 187

1. Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan 2 45° + cos 2 30° – sin 2 60

Ncert solutions class 10 chapter 8-4

First, find the values of the given trigonometric ratios

sin 30° = 1/2

cos 30° = √3/2

sin 60° = 3/2

cos 60°= 1/2

sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 =1

We know that, the values of the trigonometric ratios are:

sin 60° = √3/2

tan 45° = 1

Substitute the values in the given problem

2 tan 2 45° + cos 2 30° – sin 2 60 = 2(1) 2 + (√3/2) 2 -(√3/2) 2

2 tan 2 45° + cos 2 30° – sin 2 60 = 2 + 0

2 tan 2 45° + cos 2 30° – sin 2 60 = 2

(iii) cos 45°/(sec 30°+cosec 30°)

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

Substitute the values, we get

Ncert solutions class 10 chapter 8-5

Now, multiply both the numerator and denominator by √2 , we get

Ncert solutions class 10 chapter 8-6

Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8

Ncert solutions class 10 chapter 8-7

cosec 60° = 2/√3

cos 60° = 1/2

cot 45° = 1

Substitute the values in the given problem, we get

Ncert solutions class 10 chapter 8-8

Now, substitute the values in the given problem, we get

(5cos 2 60° + 4sec 2 30° – tan 2 45°)/(sin 2 30° + cos 2 30°)

= 5(1/2) 2 +4(2/√3) 2 -1 2 /(1/2) 2 +(√3/2) 2

  = (5/4+16/3-1)/(1/4+3/4)

= (15+64-12)/12/(4/4)

2. Choose the correct option and justify your choice : (i) 2tan 30°/1+tan 2 30° = (A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30° (ii) 1-tan 2 45°/1+tan 2 45° = (A) tan 90°            (B) 1                    (C) sin 45°            (D) 0 (iii)  sin 2A = 2 sin A is true when A = (A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1-tan 2 30° = (A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

(i) (A) is correct.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan 30°/1+tan 2 30° = 2(1/√3)/1+(1/√3) 2

= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

The obtained solution is equivalent to the trigonometric ratio sin 60°

(ii) (D) is correct.

Substitute the of tan 45° in the given equation

1-tan 2 45°/1+tan 2 45° = (1-1 2 )/(1+1 2 )

The solution of the above equation is 0.

(iii) (A) is correct.

To find the value of A, substitute the degree given in the options one by one

sin 2A = 2 sin A is true when A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

Apply the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

Now, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to cos value, i.e., cos 0 =1

Therefore, ⇒ A = 0°

(iv) (C) is correct.

2tan30°/1-tan 2 30° =  2(1/√3)/1-(1/√3) 2

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The value of the given equation is equivalent to tan 60°.

3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

Therefore A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Justification:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Since the values obtained are not equal, the solution is false.

According to the values obtained as per the unit circle, the values of sin are:

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Thus the value of sin θ increases as θ increases. Hence, the statement is true

(iii) False.

According to the values obtained as per the unit circle, the values of cos are:

cos 60° = 1/2

cos 90° = 0

Thus, the value of cos θ decreases as θ increases. So, the statement given above is false.

sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false.

Since cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

Now substitute A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Hence, it is true

Exercise 8.3 Page: 189

1. Evaluate :

(i) sin 18°/cos 72°        

(ii) tan 26°/cot 64°      

(iii)  cos 48° – sin 42°      

(iv)  cosec 31° – sec 59°

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 26°, which is equal to the cot 64°.

= tan (90° – 26°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0

trigonometry questions ncert

2.  Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

trigonometry questions ncert

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A .

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

trigonometry questions ncert

4.  If tan A = cot B, prove that A + B = 90°.

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

trigonometry questions ncert

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that

    sin (B+C/2) = cos A/2

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

Exercise 8.4 Page: 193

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas

cosec 2 A   – cot 2 A = 1

cosec 2 A = 1 + cot 2 A

Since cosec function is the inverse of sin function, it is written as

1/sin 2 A = 1 + cot 2 A

Now, rearrange the terms, it becomes

sin 2 A = 1/(1+cot 2 A)

Now, take square roots on both sides, we get

sin A = ±1/(√(1+cot 2 A)

The above equation defines the sin function in terms of cot function

Now, to express sec function in terms of cot function, use this formula

sin 2 A = 1/ (1+cot 2 A)

Now, represent the sin function as cos function

1 – cos 2 A = 1/ (1+cot 2 A)

Rearrange the terms,

cos 2 A = 1 – 1/(1+cot 2 A)

⇒cos 2 A = (1-1+cot 2 A)/(1+cot 2 A)

Since sec function is the inverse of cos function,

⇒ 1/sec 2 A = cot 2 A/(1+cot 2 A)

Take the reciprocal and square roots on both sides, we get

⇒ sec A = ±√ (1+cot 2 A)/cotA

Now, to express tan function in terms of cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since cot function is the inverse of tan function, it is rewritten as

tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:

cos 2 A + sin 2 A = 1

Rearrange the terms

sin 2 A = 1 – cos 2 A

sin 2 A = 1 – (1/sec 2 A)

sin 2 A = (sec 2 A-1)/sec 2 A

sin A = ± √(sec 2 A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec 2 A-1)

Now, tan A function in terms of sec A:

sec 2 A – tan 2 A = 1

⇒ tan 2 A = sec 2 A – 1

tan A = √(sec 2 A – 1)

cot A function in terms of sec A:

⇒ cot A = 1/tan A

cot A = ±1/√(sec 2 A – 1)

3. Evaluate:

(i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii)  sin 25° cos 65° + cos 25° sin 65°

(i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin 2 (90°-27°) + sin 2 27°] / [cos 2 (90°-73°) + cos 2 73°)]

= (cos 2 27°   + sin 2 27°)/(sin 2 27° + cos 2 73°)

= 1/1 =1                       (since sin 2 A + cos 2 A = 1)

Therefore, (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos 2 65°   + sin 2 65° = 1 (since sin 2 A + cos 2 A = 1)

Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

4. Choose the correct option. Justify your choice. (i) 9 sec 2 A – 9 tan 2 A = (A) 1                 (B) 9              (C) 8                (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0                 (B) 1              (C) 2                (D) – 1 (iii) (sec A + tan A) (1 – sin A) = (A) sec A           (B) sin A        (C) cosec A      (D) cos A

(iv) 1+tan 2 A/1+cot 2 A = 

      (A) sec 2 A                 (B) -1              (C) cot 2 A                (D) tan 2 A

(i) (B) is correct.

Take 9 outside, and it becomes

9 sec 2 A – 9 tan 2 A

= 9 (sec 2 A – tan 2 A)

= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

Therefore, 9 sec 2 A – 9 tan 2 A = 9

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

We know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

Now, substitute the above values in the given problem, we get

= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)

Simplify the above equation,

= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ) 2 -1 2 /(cos θ sin θ)

= (cos 2 θ + sin 2 θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos 2 θ + sin 2 θ = 1)

= (2cos θ sin θ)/(cos θ sin θ) = 2

Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2

(iii) (D) is correct.

Sec A= 1/cos A

Tan A = sin A / cos A

(secA + tanA) (1 – sinA)

= (1/cos A + sin A/cos A) (1 – sinA)

= (1+sin A/cos A) (1 – sinA)

= (1 – sin 2 A)/cos A

= cos 2 A/cos A = cos A

Therefore, (secA + tanA) (1 – sinA) = cos A

(iv) (D) is correct.

tan 2 A =1/cot 2 A

Now, substitute this in the given problem, we get

1+tan 2 A/1+cot 2 A

= (1+1/cot 2 A)/1+cot 2 A

= (cot 2 A+1/cot 2 A)×(1/1+cot 2 A)

= 1/cot 2 A = tan 2 A

So, 1+tan 2 A/1+cot 2 A = tan 2 A

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ – cot θ) 2  = (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

     [Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin 2 A/(1-cos A)  

     [Hint : Simplify LHS and RHS separately]

(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec 2 A = 1+cot 2 A.

Ncert solutions class 10 chapter 8-10

(vii) (sin θ – 2sin 3 θ)/(2cos 3 θ-cos θ) = tan θ (viii) (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) [Hint : Simplify LHS and RHS separately] (x) (1+tan 2 A/1+cot 2 A) = (1-tan A/1-cot A) 2  =   tan 2 A

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ) 2

The above equation is in the form of (a-b) 2 , and expand it

Since (a-b) 2 = a 2 + b 2 – 2ab

Here a = cosec θ and b = cot θ

= (cosec 2 θ + cot 2 θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin 2 θ + cos 2 θ/sin 2 θ – 2cos θ/sin 2 θ)

= (1 + cos 2 θ – 2cos θ)/(1 – cos 2 θ)

= (1-cos θ) 2 /(1 – cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) = R.H.S.

Therefore, (cosec θ – cot θ) 2  = (1-cos θ)/(1+cos θ)

(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Now, take the L.H.S of the given equation.

L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)

= [cos 2 A + (1+sin A) 2 ]/(1+sin A)cos A

= (cos 2 A + sin 2 A + 1 + 2sin A)/(1+sin A) cos A

Since cos 2 A + sin 2 A = 1, we can write it as

= (1 + 1 + 2sin A)/(1+sin A) cos A

= (2+ 2sin A)/(1+sin A)cos A

= 2(1+sin A)/(1+sin A)cos A

= 2/cos A = 2 sec A = R.H.S.

L.H.S. = R.H.S.

(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)

We know that tan θ =sin θ/cos θ

Now, substitute it in the given equation, to convert it in a simplified form

= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]

= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]

= sin 2 θ/[cos θ(sin θ-cos θ)] + cos 2 θ/[sin θ(cos θ-sin θ)]

= sin 2 θ/[cos θ(sin θ-cos θ)] – cos 2 θ/[sin θ(sin θ-cos θ)]

= 1/(sin θ-cos θ) [(sin 2 θ/cos θ) – (cos 2 θ/sin θ)]

= 1/(sin θ-cos θ) × [(sin 3 θ – cos 3 θ)/sin θ cos θ]

= [(sin θ-cos θ)(sin 2 θ+cos 2 θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + 1

= 1 + sec θ cosec θ = R.H.S.

Therefore, L.H.S. = R.H.S.

Hence proved

(iv)  (1 + sec A)/sec A = sin 2 A/(1-cos A)

First find the simplified form of L.H.S

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin 2 A/(1-cos A)

We know that sin 2 A = (1 – cos 2 A), we get

= (1 – cos 2 A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin 2 A/(1-cos A)= cos A + 1

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec 2 A = 1+cot 2 A.

With the help of identity function, cosec 2 A = 1+cot 2 A, let us prove the above equation.

L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)

Divide the numerator and denominator by sin A, we get

= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A

We know that cos A/sin A = cot A and 1/sin A = cosec A

= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)

= (cot A – cosec 2 A + cot 2 A + cosec A)/(cot A+ 1 – cosec A) (using cosec 2 A – cot 2 A = 1

= [(cot A + cosec A) – (cosec 2 A – cot 2 A)]/(cot A+ 1 – cosec A)

= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)

=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)

=  cot A + cosec A = R.H.S.

Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A

Hence Proved

trigonometry questions ncert

First divide the numerator and denominator of L.H.S. by cos A,

Ncert solutions class 10 chapter 8-12

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= √(sec A+ tan A)/(sec A-tan A)

Now using rationalization, we get

Ncert solutions class 10 chapter 8-13

= (sec A + tan A)/1

= sec A + tan A = R.H.S

(vii) (sin θ – 2sin 3 θ)/(2cos 3 θ-cos θ) = tan θ

L.H.S. = (sin θ – 2sin 3 θ)/(2cos 3 θ – cos θ)

Take sin θ as in numerator and cos θ in denominator as outside, it becomes

= [sin θ(1 – 2sin 2 θ)]/[cos θ(2cos 2 θ- 1)]

We know that sin 2 θ = 1-cos 2 θ

= sin θ[1 – 2(1-cos 2 θ)]/[cos θ(2cos 2 θ -1)]

= [sin θ(2cos 2 θ -1)]/[cos θ(2cos 2 θ -1)]

= tan θ = R.H.S.

(viii) (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A

L.H.S. = (sin A + cosec A) 2  + (cos A + sec A) 2

It is of the form (a+b) 2 , expand it

(a+b) 2 =a 2 + b 2 +2ab

= (sin 2 A + cosec 2 A + 2 sin A cosec A) + (cos 2 A + sec 2 A + 2 cos A sec A)

= (sin 2 A + cos 2 A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan 2 A + 1 + cot 2 A

= 1 + 2 + 2 + 2 + tan 2 A + cot 2 A

= 7+tan 2 A+cot 2 A = R.H.S.

Therefore, (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A

trigonometry questions ncert

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

First, find the simplified form of L.H.S

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (1/sin A – sin A)(1/cos A – cos A)

= [(1-sin 2 A)/sin A][(1-cos 2 A)/cos A]

= (cos 2 A/sin A)×(sin 2 A/cos A)

= cos A sin A

Now, simplify the R.H.S

R.H.S. = 1/(tan A+cotA)

= 1/(sin A/cos A +cos A/sin A)

= 1/[(sin 2 A+cos 2 A)/sin A cos A]

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

(x)  (1+tan 2 A/1+cot 2 A) = (1-tan A/1-cot A) 2  =   tan 2 A

L.H.S. = (1+tan 2 A/1+cot 2 A)

Since cot function is the inverse of tan function,

= (1+tan 2 A/1+1/tan 2 A)

= 1+tan 2 A/[(1+tan 2 A)/tan 2 A]

Now cancel the 1+tan 2 A terms, we get

(1+tan 2 A/1+cot 2 A) = tan 2 A

(1-tan A/1-cot A) 2  =   tan 2 A

NCERT Solutions for Class 10 Chapter 8 – Introduction to Trigonometry

For the  Class 10 CBSE Maths examinations, out of the 80 marks (combined), 12 marks are assigned from the unit 5 “Trigonometry”. The paper consists of 4 parts. Each part carries different marks and the questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. You can expect at least 2-3 compulsory questions from this chapter. The main topics covered in this chapter include:

8.1 Introduction

You have already studied about triangles, and in particular, right triangles, in your earlier classes. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We also define the trigonometric ratios for angles of measure 0 o and 90 o . We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Identities

You have studied the concept of ratio, in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The topic is explained with suitable examples by using different functions of Trigonometry.

8.3 Trigonometric Ratios of Some Specific Angles

From Geometry, you are already familiar with the construction of angles of 30 0 , 45 o , 60 o and 90 0 . In this section, we will find the values of the trigonometric ratios for these angles and for 0 o . It explains Trigonometric Ratios of 45 o , Trigonometric Ratios of 30 o and 60 0 , Trigonometric Ratios of 0 o and 90 o with suitable examples.

8.4 Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum equals 90 o . The topic discusses various formulas to solve numerical problems related to trigonometric ratios.

8.5 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true of all values of the angles involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

8.6 Summary

Summary is the brief of concepts that includes all the important points which you need to memorize to solve numerical problems related to the chapter.

List of Exercises in Class 10 Maths Chapter 8 :

Exercise 8.1 Solutions – 11 Questions (7 short answers, 3 long answers, 1 short answer with reasoning)

Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions – 7 Questions (5 short answers, 2 long answers)

Exercise 8.4 Solutions – 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

Hence, these  NCERT Solutions for Class 10 Maths  will help students understand different types of questions and their answers along with key shortcuts and diagrammatic representations. All the NCERT Solutions for Class 10 Maths Chapter 8   PDF given here are presented in simple language. Comprehending these solutions thoroughly will aid students to solve complex problems effortlessly.

The faculty have curated the NCERT Solutions for Class 10 in a lucid manner to improve the problem-solving abilities among the students. For a more clear idea about Introduction To Trigonometry, students can refer to the study materials available at BYJU’S.

  • RD Sharma Solutions for Class 10 Maths Introduction To Trigonometry

NCERT Solution for class 10 Maths chapter 8

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 8

List out the frequently-asked topics of chapter 8 of ncert solutions in the cbse exam of class 10 maths., why should we download ncert solutions for class 10 maths chapter 8 from byju’s, is ncert solutions for class 10 maths chapter 8 important from the exam point of view, leave a comment cancel reply.

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trigonometry questions ncert

I can’t understand how 6/4√3=√3/2

we have, 6/4√3=√3/2 6/4√3= 3/2√3 (√3 * √3)/2√3 Since,√3/√3 = 1 we get the answer, √3/2

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

trigonometry questions ncert

Get here the NCERT Solutions for class 10 Maths Chapter 8 Introduction to Trigonometry in English and Hindi Medium for session 2024-25. The solution of chapter 8 class 10th Maths is modified and revised as per the new textbooks published by NCERT for academic year 2024-25.

Class 10 Maths Chapter 8 Exercise 8.1 Solutions

  • Class 10th Maths Exercise 8.1 in English
  • Class 10th Maths Exercise 8.1 in Hindi

Class 10 Maths Chapter 8 Exercise 8.2 Solutions

  • Class 10th Maths Exercise 8.2 in English
  • Class 10th Maths Exercise 8.2 in Hindi

Class 10 Maths Chapter 8 Exercise 8.3 Solutions

  • Class 10th Maths Exercise 8.3 in English
  • Class 10th Maths Exercise 8.3 in Hindi

10th Maths Chapter 8 Solutions for State Boards

  • Class 10 Maths Chapter 8 Exercise 8.1
  • Class 10 Maths Chapter 8 Exercise 8.2
  • Class 10 Maths Chapter 8 Exercise 8.3
  • Class 10 Maths Chapter 8 Exercise 8.4
  • Class 10 Maths Solutions Page
  • Class 10th all Subjects NCERT Solutions

All the contents are updated for new academic session 2024-25 based on latest NCERT Books. UP Board students are also using NCERT Textbooks for their exams now. So, they also can download UP Board Solutions for Class 10 Maths Chapter 8 Solutions in Hindi Medium from this page. Solutions are given in videos format also describing in English and Hindi.

Class 10 Maths Chapter 8 Topics

NCERT Solutions and Offline apps for UP Board and CBSE Board are also given free to download. Download Offline as well as Online Apps based on updated NCERT Solutions based on latest NCERT books 2024-25. Revision books are also available to download. Description, history and identities related to Trigonometry is given below.

NCERT Solutions for class 10 Maths Chapter 8 Introduction to Trigonometry exercises from 8.1 to 8.3 are given to use online of download in PDF format free. These solutions are in Hindi and English Medium format. If you have any doubt, please visit to Discussion Forum to ask your doubts.

Class 10 Maths Chapter 8 Solutions

1. The creator of trigonometry is said to have been the Greek Mathematician Hipparchus of the 2nd century BC. 2. The word Trigonometry which means triangle measurement is credited to Bastholoman Pitiscus (1561-1613). 3. The first use of the idea of ‘sine’ can be found in the work of ‘Aryabhatiyam’ of Aryabhata in 500 AD. Aryabhata used the word Ardha-jya for the half-chord, which was shortened to Jya or Jiva in due course. When the Aryabhatiyam was translated into Arabic, the word Jiva was retained. It was further translated into ‘Sinus’, which means curve in Latin. The word ‘Sinus’ also used as sine was first abbreviated and used as ‘sin’ by an English professor of astronomy Edmund Gunter (1581-1626). 4. The origin of the terms ‘Cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhata called it Kotijya. The name cosinus originated with Edmund Gunter. In 1674, another English mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

Trigonometry - Trigonometric Identities

Trigonometry - Trigonometric Ratios

How many exercise wise questions are there in class 10 Maths chapter 8?

There are 4 exercises in class 10 math chapter 8 Introduction to Trigonometry. In first exercise (Ex 8.1), there are in all 11 questions. In second exercise (Ex 8.2), there are in all 4 questions. In third exercise (Ex 8.3), there are 5 questions. So, there are in all 20 questions in class 10 math chapter 8 Introduction to Trigonometry. There are in all 15 examples in class 10 math chapter 8 Introduction to Trigonometry. Examples 1, 2, 3, 4, 5 are based on Ex 8.1. Examples 6, 7, 8 are based on Ex 8.2. Examples 12, 13, 14, 15 are based on Ex 8.3.

What are the important examples from chapter 8 class 10 math?

Examples given in NCERT books are always important for CBSE exams. Examples 3, 4, 5, 7, 8, 10, 11, 13, 14 and 15 of chapter 8 (Introduction to Trigonometry) of class 10 math are the important examples from exam point of view. In exam questions can come from these examples.

Which are the main topics students go through in chapter 8 of class 10th Mathematics?

In chapter 8 Introduction to Trigonometry of class 10th math, Students will study: 1) Trigonometric Ratios. 2) Trigonometric Ratios of Some Specific Angles. 3) Trigonometric Identities. This chapter is very interesting. Also this chapter is completely new for students.

Why is chapter 8 of Class 10th Maths important?

Chapter 8 (Introduction to Trigonometry) of 10th Math is important because there is a chapter in class 11th math named Trigonometric functions and chapter 8 (Introduction to Trigonometry) of 10th Math works as a base for that chapter. Also from exam point of view chapter 8 (Introduction to Trigonometry) of 10th Math is very important. Every year 7-8 marks questions come from this chapter.

What do you learn about Trigonometry in Class 10 Maths Chapter 8?

Trigonometry is the oldest branch of mathematics. This concept was first used by Aryabhata in Aryabhatiyam in 500 A.D. Trigonometry is a word consisting of three Greek words: Tri-Gon-Metron. ‘Tri’ means three, ‘Gon’ means side and ‘Metron’ means measure. So, trigonometry is the study related to the measure of sides and angles of a triangle in particular, right triangles (in CBSE class 10).

How can I score more that 90% in class 10 Maths Chpater 8 Trigonometry?

To score well in trigonometry, we should practice hard the entire chapter 8 in 10th Maths. Specially exercise 8.4 needs more time to spend for solving the questions based on identities. All the parts of question 5 of exercise 8.4 asked in exams frequently.

Is there any application of trigo as a tool in astronomy for Class 10 Mathematics 8th Chapter?

Trigonometry is used in astronomy to determine the position and the path of celestial objects. Astronomers use it to find the distance of the stars and planets from the Earth. Captain of a ship uses it to find the direction and the distance of islands and light houses from the sea. Surveyors use to map the new lands.

What is the objective of Chapter 8 in Class 10 Maths Trigonometry?

Objective of Class 10 Trigonometry: Identifying the opposite side, adjacent side and hypotenuse of right triangle with respect to given angle A. Defining the six rations (sine, cosine, tangent, secant. cosecant and cotangent) related to the sides of a right angled triangle. Finding the values of trigonometric rations of a given right angled triangle. Finding the values of trigonometry rations of some standard angles (0, 30, 45, 60 and 90) in degrees. Using complementary angles and applying it into trigonometric identities to prove another identities.

« Chapter 7: Coordinate Geometry

Chapter 9: some applications of trigonometry ».

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NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are formulated to explain the fundamental application of trigonometric formulas and the graphs of their functions. Trigonometry is an important part of Class 11 maths that involves studying various relations between sides and angles of triangles . Trigonometric Functions are applied to define these relations and have numerous applications in various other fields, including science and engineering. They are often used for basic geometric calculations and to explain numeric solutions. The knowledge of Trigonometric Functions is vital for subjects like astronomy and geography. NCERT Solutions Class 11 Maths Chapter 3 will offer the right foundational skills in students to study trigonometric relations and functions along with their practical applications.

Chapter 3 of Class 11 Maths will enable students to generalize the concept of trigonometric ratios to trigonometric functions. Learning about the properties of Trigonometric Functions and operations based on them is crucial for math studies. With the regular practice of Class 11 Maths NCERT Solutions Chapter 3, students will quickly master the solutions to equations using Trigonometric Functions. The wide range of problems and examples provided in these solutions are beneficial in promoting an in-depth understanding of concepts. To learn and practice with NCERT Solutions Chapter 3 Trigonometric Functions, download the exercises provided in the links below.

  • NCERT Solutions Class 11 Maths Chapter 3 Ex 3.1
  • NCERT Solutions Class 11 Maths Chapter 3 Ex 3.2
  • NCERT Solutions Class 11 Maths Chapter 3 Ex 3.3
  • NCERT Solutions Class 11 Maths Chapter 3 Ex 3.4
  • NCERT Solutions Class 11 Maths Chapter 3 Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 3 PDF

Trigonometry sees the use of many formulas, theorems, and steps to solve the sums hence, ensuring that kids allot an ample amount of time for practice is very important. NCERT Solutions for Class 11 Maths Chapter 3 are proficiently modeled to support higher-level math learning. The comprehensive format of these solutions is highly reliable to enhance problem-solving skills in engaging ways. To learn and practice trigonometric functions with these solutions, click on the links of the pdf file given below.

☛ Download Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions

NCERT Class 11 Maths Chapter 3   Download PDF

NCERT Solutions Class 11 Maths Chapter 3 1

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions are extremely beneficial in developing mathematical reasoning in students. With the help of these well-structured resources, students will gain a simplistic approach for efficient exam preparation. The carefully placed illustrations, questions, and examples present in these solutions are sufficient to enhance the fundamental math knowledge to excel in exams. Kids can easily plan their curriculum and revision format with the help of these solutions. To practice the exercise-wise NCERT Solutions Class 11 Maths Trigonometric Functions, try the links given below.

  • Class 11 Maths Chapter 3 Ex 3.1 - 7 Questions
  • Class 11 Maths Chapter 3 Ex 3.2 - 10 Questions
  • Class 11 Maths Chapter 3 Ex 3.3 - 25 Questions
  • Class 11 Maths Chapter 3 Ex 3.4 - 9 Questions
  • Class 11 Maths Chapter 3 Miscellaneous Ex - 10 Questions

☛  Download Class 11 Maths Chapter 3 NCERT Book

Topics Covered: NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions cover some important topics, including an introduction to trigonometric ratios and identities, the measure of angles, signs, domain and range of trigonometric functions . The other important topic included in these solutions is the trigonometric solutions of the sum and difference of two angles using formulas .

Total Questions: Class 11 Maths Chapter 3 Trigonometric Functions has 51 questions in 4 exercises plus 10 questions in one miscellaneous exercise. These are primarily based on the representation of trigonometric functions, their applications, and formulas.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 3

Memorizing important formulas and concepts is necessary to understand the terms and operations related to trigonometric functions clearly. NCERT Solutions Class 11 Maths Chapter 3 will provide detailed knowledge of all these along with their applications through interesting illustrations. Each concept in these solutions is well-explained with suitable examples and sample problems to promote an in-depth understanding of this topic. Formulas form an integral part of this lesson hence, creating a formulas sheet can be useful for students when they need to practice this topic. Some of the important terms, formulas, and concepts related to Trigonometric Functions explained in these solutions are listed below:

  • Trigonometric Functions: Trigonometric Functions are real functions that relate the angle of a r ight-angled triangle to the ratio of the length of its sides. Sin, Cos, and Tan are the three primary functions.
  • Trigonometric Equations: Trigonometric equations are equations that involve the use of trigonometric functions. These trigonometric equations are also known as trigonometric identities when the values of unknown angles for which the functions are defined are satisfied.
  • Trigonometric Identities: A trigonometric equation that involves the sum and difference of angles represent a trigonometric identity. For example, sin 2 θ + cos 2 θ = 1
  • Sum and Difference Identities: The sum and difference identities include the following formulas.
  • sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
  • cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
  • tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
  • sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
  • cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
  • tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

FAQs on NCERT Solutions Class 11 Maths Chapter 3

What is the importance of ncert solutions for class 11 maths chapter 3 trigonometric functions.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are designed by well-qualified teachers and math experts to promote proficient math learning. Each exercise of these solutions is formulated as per the CBSE guidelines to support in-depth learning of Trigonometric Functions and their relations. These competently structured solutions are suitable to enhance math proficiency in students. Regular practice of these solutions will boost students’ confidence in scoring well.

What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 briefly introduces trigonometric ratios and identities along with some core concepts previously studied in grade 10. The important topics covered in these solutions are angle measures, trigonometric functions, their signs, domain, and range. The sum and difference of two angles using trigonometric functions are also included in this chapter.

Do I Need to Practice all Questions Provided in Class 11 Maths NCERT Solutions Trigonometric Functions?

NCERT Solutions Class 11 Maths Trigonometric Functions are designed in an effective way to facilitate the in-depth learning of concepts. Every question included in these solutions is aimed to improve the conceptual clarity for students to master them easily. The fundamental knowledge of Trigonometric Functions and their use will allow students to apply their knowledge in practical situations. These solutions also form the basis for studying advanced math topics, including calculus. Thus, all sums must be practiced with laser focus.

How Many Questions are there in Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions has 61 questions in 5 exercises. These problems are sufficient to provide a deep-seated understanding of the entire topic of trigonometric functions, including important formulas, identities, ratios, and operations related to Trigonometric Functions. The jargon-free and lucid math vocabulary used in these solutions are quite proficient in easily imparting a clear step-by-step understanding of each topic as well as subtopics.

What are the Important Formulas in Class 11 Maths NCERT Solutions Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 explains the formulas related to the sum and difference of two angles in trigonometric functions. Some of the important concepts related to this topic are based on deriving expressions for trigonometric identities of the sum and difference of angles. These vital concepts are described comprehensively with the help of interesting examples for students to grasp them better. Additionally, kids need to also revise the formulas they have encountered in previous classes as those are also necessary to attempt class 11 trigonometry problems.

Why Should I Practice NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is a reliable source of learning that gives full guidance for excellent exam preparation. With the help of these solutions, students can cover the whole syllabus of CBSE Class 11 Maths Chapter 3. The format of these solutions is quite intuitive to promote simple and easy learning. Thus, to get the best results kids need to practice these solutions.

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

  • What is a positive or a negative angle
  • Measuring angles in Degree , Minutes and Seconds
  • Radian measure of an angle
  • Converting Degree to Radians , and vice-versa
  • Sign of sin, cos, tan in all 4 quadrants
  • Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
  • Finding Value of trigonometric functions, given angle
  • Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
  • Finding principal and general solutions of a trigonometric equation
  • Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions

Ncert solutions for class 11 maths chapter 3 – trigonometric functions pdf.

Free PDF of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 3 – Trigonometric Functions Maths NCERT Solutions for Class 11 to help you to score more marks in your board exams and as well as competitive exams.

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NCERT Solutions for Class 11 Maths Chapter 3 provides 100% accurate and comprehensive answers to all questions from NCERT textbooks. All students aspiring to excel in their entrance exams should refer to these study guides for more profound knowledge and better grades in this subject. These solutions have been drafted by teaching experts following the latest CBSE guidelines.

The other chapters in class 11 maths ncert solutions are very important and available in PDF format. You can also download physics class 11 and chemistry class 11 solutions on vedantu site.

Trigonometric Functions Chapter at a Glance - Class 11 NCERT Solutions

$1^a-\left(\frac{1}{360}\right)^2$ of a revolution

$1^{\circ}-60^{\prime}, 1^{\prime}-60^{\circ}$

1 Revolution $-2 \pi$ radians

$1^2=\frac{\pi}{180}$ radian

If in a circle of radius $r$, an arc of length $l$ subtends an angle of $\theta$ radians, then $\theta^*=\frac{l}{r}$

Trigonometric identities

(i) $\sin ^2 \theta+\cos ^2 \theta=1$

(ii) $\sec ^2 \theta-1+\tan ^2 \theta$

(iii) $\operatorname{coses}^2 \theta=1+\cot ^2 \theta$

(i) $\sin (-x)=-\sin x$

(ii) $\cos (-x)=\cos x$

(iii) $\tan (-x)=-\tan x$

(iv) $\operatorname{cosec}(-x)=-\operatorname{cosec} x$

(v) $\sec (-x)-\sec x$

(vi) $\cot (-x)=-\cot x$

(i) $\sin \left(\frac{\pi}{2} \pm \theta\right)=\cos \theta$

(ii) $\cos \left(\frac{\pi}{2} \pm \theta\right)=\mp \sin \theta$

(iii) $\tan \left(\frac{\pi}{2} \pm \theta\right)=\mp \cot \theta$

(iv) $\cot \left(\frac{\pi}{2} \pm \theta\right)=\mp \tan \theta$

(v) $\sec \left(\frac{\pi}{2} \pm \theta\right)=\mp \operatorname{cosec} \theta$

(vi) $\operatorname{cosec}\left(\frac{\pi}{2} \pm \theta\right)=\sec \theta$

(i) $\sin (\pi \pm \theta)= \pm \sin \theta$

(ii) $\cos (\pi \pm \theta)=-\cos \theta$

(iii) $\tan (\pi \pm \theta)=\mp \tan \theta$

(iv) $\cot (\pi \pm \theta)=\mp \cot \theta$

(v) $\sec (\pi \pm \theta)=-\sec \theta$

(vi) $\operatorname{coses}(\pi \pm \theta)=\mp \operatorname{cosec} \theta$

(i) $\sin \left(\frac{3 \pi}{2} \pm \theta\right)=-\cos \theta$

(ii) $\cos \left(\frac{3 \pi}{2} \pm \theta\right)=\sin \theta$

(iii) $\tan \left(\frac{3 \pi}{2} \pm \theta\right)=-\cot \theta$

(iv) $\cot \left(\frac{3 \pi}{2} \pm \theta\right)=\tan \theta$

(v) $\sec \left(\frac{3 \pi}{2} \pm \theta\right)-\operatorname{cosec} \theta$

(vi) $\operatorname{cosec}\left(\frac{3 \pi}{2} \pm \theta\right)=-\sec \theta$

(i) $\sin (2 \pi \pm \theta)=\mp \sin \theta$

(ii) $\cos (2 \pi \pm \theta)=\cos \theta$

(iii) $\tan (2 \pi \pm \theta)=\mp \tan \theta$

(iv) $\cot (2 \pi \pm \theta)=\mp \cot \theta$

(v) $\sec (2 \pi \pm \theta)=\sec \theta$

(vi) $\operatorname{cosec}(2 \pi \pm \theta)-\mp \operatorname{cosec} \theta$

Trigonometric functions of sum and difference of angles:

$\sin (x \pm y)=\sin x \cos y \pm \cos x \sin y$

$\cos (x \pm y)=\cos x \cos y \mp \sin x \sin y$

If nome of the angles $x, y$ and $(x \pm y)$ is an odd multiple of $\frac{\pi}{2}$, then $\tan (x \pm y)=\frac{\tan x \pm \tan y}{17 \tan x \tan y}$

If nome of the angles $x, y$ and $(x \pm y)$ is a multiple of $\pi$, then $\cot (x \pm y)-\frac{\cot x \cot y \mp 1}{\cot y \pm \cot x}$

Multiple and Submultiple Angle formula:

$\cos 2 x=\cos ^2 x-\sin ^2 x$

$$ \begin{aligned} & -2 \cos ^2 x-1-1-2 \sin ^2 x \\ & =\frac{1-\tan ^2 x}{1+\tan ^2 x} \end{aligned} $$

$\sin 2 x=2 \sin x \cdot \cos x=\frac{2 \tan x}{1+\tan ^2 x}$

$\tan 2 x=\frac{2 \tan x}{1-\tan ^2 x}$

$\sin 3 x=3 \sin x-4 \sin ^3 x$

$\cos 3 x-4 \cos ^3 x-3 \cos x$

$\tan 3 x=\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}$

Trigonometric Transformation formulae:

$\sin C+\sin D$ $$ =2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) $$ $\sin C-\sin D$ $$ =2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) $$ $\cos C+\cos D$ $$ 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) $$ $\cos C-\cos D$ $$ =-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) $$ - $2 \sin A \cos B$ $$ -\sin (A+B)+\sin (A-B) $$ - $2 \cos A \sin B$ $$ -\sin (A+B)-\sin (A-B) $$ - $2 \cos A \cos B$ $$ =\cos (A+B)+\cos (A-B) $$ - $2 \sin A \sin B$ $$ -\cos (A-B)-\cos (A+B) $$

(i) $\sin \theta=\sin \alpha$ $$ \Rightarrow \theta=n \pi+(-1)^n \alpha, n \in Z $$

(ii)$\cos \theta=\cos x$ $$ \Rightarrow \theta=2 n x \pm \alpha, n \in Z $$

(iii)$\tan \theta=\tan x$ $$ \Rightarrow \theta=n \pi+\alpha, n \in Z $$ $\sin ^2 \theta=\sin ^2 \alpha$

(iv) $\left.\cos ^2 \theta=\cos ^2 \alpha\right\} \theta=n \pi \pm a, n \in Z$ $\tan ^2 x=\tan ^2 \alpha$

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Exercises under NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Exercise 3.1: In this exercise, students are introduced to trigonometric ratios of acute angles and their applications in solving problems related to heights and distances. The exercise covers basic concepts such as the definition of trigonometric ratios, Pythagoras theorem, and the concept of complementary angles.

Exercise 3.2: This exercise focuses on the evaluation of trigonometric ratios of special angles such as 0, 30, 45, 60, and 90 degrees. The exercise also includes the derivation of trigonometric ratios of 30, 45, and 60 degrees using the concept of the unit circle.

Exercise 3.3: This exercise covers the trigonometric ratios of angles between 0 and 90 degrees. The exercise includes problems on finding the values of trigonometric ratios using various techniques such as the use of Pythagoras theorem, the double-angle formula, and the half-angle formula.

Exercise 3.4: In this exercise, students learn about the trigonometric ratios of complementary angles and the concept of co-functions. The exercise covers problems on finding the values of trigonometric ratios using the co-function identity.

Miscellaneous Exercise: This exercise includes a variety of problems that cover different topics such as the use of trigonometric ratios in solving real-life problems, the use of trigonometric identities to simplify expressions, and the solution of trigonometric equations. This exercise provides an opportunity for students to apply the concepts learned in the chapter to solve more complex problems.

Access NCERT Solution for Class 11 Maths Chapter 3 - Trigonometric Functions

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) $\text{2}{{\text{5}}^{\text{o}}}$

Ans: We know that $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian 

hence, 

$\text{2}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 25}$ radian  

      $\text{=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{36}}$radian

(ii) $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ }$

Ans: Here we have,

  $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-47}{{\dfrac{\text{1}}{\text{2}}}^{\text{o}}}$   

           $\text{=-}\dfrac{\text{95}}{\text{2}}$ degree

Since we know that, $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{\text{1}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}$ radian 

$\text{-}\dfrac{\text{95}}{\text{2}}$ degree$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{-95}}{\text{2}} \right)$ radian

                  $\text{=}\left( \dfrac{\text{-19}}{\text{36 }\!\!\times\!\!\text{ 2}} \right)\text{ }\!\!\pi\!\!\text{ }$  radian

                  $\text{=}\dfrac{\text{-19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$radian

$\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-}\dfrac{\text{19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$ radian

(iii) $\text{24}{{\text{0}}^{\text{o}}}$

Ans: We know that,

$\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

$\text{24}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 240}$ radian

       $\text{=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$radian

(iv) $\text{52}{{\text{0}}^{\text{o}}}$

$\text{52}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 520}$ radian

       $\text{=}\dfrac{\text{26 }\!\!\pi\!\!\text{ }}{\text{9}}$radian

2. Find the degree measures corresponding to the following radian measures

(Use$\text{ }\!\!\pi\!\!\text{ =}\dfrac{\text{22}}{\text{7}}$ ) 

(i) $\dfrac{\text{11}}{\text{16}}$

 $\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{11}}{\text{16}}$ degree

              $\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11}}{\text{ }\!\!\pi\!\!\text{  }\!\!\times\!\!\text{ 4}}$degree

              \[\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11 }\!\!\times\!\!\text{ 7}}{\text{22 }\!\!\times\!\!\text{ 4}}\] degree

              $\text{=}\dfrac{\text{315}}{\text{8}}$degree

Further computing,

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=39}\dfrac{\text{3}}{\text{8}}$ degree

                $\text{=3}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{3 }\!\!\times\!\!\text{ 60}}{\text{8}}$ minutes

Since ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$

    $\dfrac{\text{11}}{\text{16}}$ radian $\text{=3}{{\text{9}}^{\text{o}}}\text{+22 }\!\!'\!\!\text{ +}\dfrac{\text{1}}{\text{2}}$minutes

Since  $\text{1 }\!\!'\!\!\text{ =60''}$ 

$\dfrac{\text{11}}{\text{16}}$ radian$\text{ }\!\!~\!\!\text{ =3}{{\text{9}}^{\text{o}}}\text{22 }\!\!'\!\!\text{ 30''}$

(ii) $\text{-4}$

$\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

$\text{-4}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\left( \text{-4} \right)$ degree

                $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7}\left( \text{-4} \right)}{\text{22}}$degree

                $\text{=}\dfrac{\text{-2520}}{\text{11}}$ degree

                $\text{=-229}\dfrac{\text{1}}{\text{11}}$degree

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{1 }\!\!\times\!\!\text{ 60}}{\text{11}}$ minutes                        

               $\text{=-22}{{\text{9}}^{\text{o}}}\text{+5 }\!\!'\!\!\text{ +}\dfrac{\text{5}}{\text{11}}$ minutes

Since $\text{1 }\!\!'\!\!\text{ =60 }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{5 }\!\!'\!\!\text{ 27''}$

(iii) $\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$

$\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ degree

                 $\text{=30}{{\text{0}}^{\text{o}}}$

(iv)$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

 $\pi $ radian$\text{=18}{{\text{0}}^{\text{o}}}$

$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

                 $\text{=21}{{\text{0}}^{\text{o}}}$

3. A wheel makes $\text{360}$ revolutions in one minute. Through how many radians does it turn in one second?

Ans: Number of revolutions the wheel makes in $\text{1}$ minute$\text{=360}$

 Number of revolutions the wheel make in $\text{1}$ second$\text{=}\dfrac{\text{360}}{\text{60}}$

                                                                                   $\text{=6}$

In one complete revolution, the wheel turns an angle of \[\text{2 }\!\!\pi\!\!\text{ }\] radian.

Hence, it will turn an angle of $\text{6 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ =12 }\!\!\pi\!\!\text{ }$ radian, in $\text{6}$ complete revolutions.

Therefore, the wheel turns an angle of $\text{12 }\!\!\pi\!\!\text{ }$ radian in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius $\text{100}$cm by an arc of length  $\text{22}$  cm.

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$                                ……(1)

Therefore, 

Substituting $\text{r=100cm}$ ,\[\text{l=22cm}\] in the formula (1) , we have,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{22}}{\text{100}}$ radian

Since $\text{1 radian=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}$ 

 $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{22}}{\text{100}}$ degree

     $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7 }\!\!\times\!\!\text{ 22}}{\text{22 }\!\!\times\!\!\text{ 100}}$degree

     $\text{=}\dfrac{\text{63}}{\text{5}}$ degree

    $\text{=12}\dfrac{\text{3}}{\text{5}}$degree

Since  ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$ , we have,

$\text{ }\!\!\theta\!\!\text{ =1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$

Hence , the required angle is $\text{1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$.

5. In a circle of diameter $\text{40}$ cm, the length of a chord is $\text{20}$ cm. Find the length of minor arc of the chord.

Ans: Given that, diameter of the circle$=40$ cm

 Hence Radius $\left( r \right)$ of the circle$=\dfrac{40}{2}cm$

                                                   $=20cm$

Let $\text{AB}$ be a chord  of the circle whose length is $20$ cm.

(Image will be uploaded soon)

             

In $\Delta \text{OAB,}$

$\text{OA=OB}$ 

     $=$ Radius of the circle

     $\text{=20cm}$

Now also, $\text{AB=20cm}$

Therefore, $\Delta \text{OAB}$ is an equilateral triangle.

\[\therefore \text{ }\!\!\theta\!\!\text{ =6}{{\text{0}}^{\text{o}}}\]

     $\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian

We know that,

Substituting $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  in the formula (1),

       $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{arc AB}}{\text{20}}$

$\text{arc AB=}\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$

Therefore, the length of the minor arc of the chord is $\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$ .

6. If in two circles, arcs of the same length subtend angles $\text{6}{{\text{0}}^{\text{o}}}$ and $\text{7}{{\text{5}}^{\text{o}}}$ at the centre, find the ratio of their radii.

Ans: Let the radii of the two circles be ${{\text{r}}_{\text{1}}}$ and ${{\text{r}}_{\text{2}}}$ . Let an arc of length ${{\text{l}}_{\text{1}}}$ subtends an angle of $\text{6}{{\text{0}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{1}}}$ , whereas let an arc of length ${{\text{l}}_{\text{2}}}$ subtends an angle of $\text{7}{{\text{5}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{2}}}$ .

Now, we have,

$\text{6}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian  and 

${{75}^{\circ }}=\dfrac{5\pi }{12}$ radian

\[\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}\]  

$\text{l=r }\!\!\theta\!\!\text{ }$          

Hence we obtain,                   

$\text{l=}\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

and  $\text{l=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

according to the ${{\text{l}}_{\text{1}}}\text{=}{{\text{l}}_{\text{2}}}$ 

thus we have,

$\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3 }\!\!\pi\!\!\text{ }}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

  ${{\text{r}}_{\text{1}}}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5}}{\text{4}}$

 $\dfrac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}}\text{=}\dfrac{\text{5}}{\text{4}}$

Hence , the ratio of the radii is $\text{5:4}$ .

7. Find the angle in radian through which a pendulum swings if its length is $\text{75}$ cm and the tip describes an arc of length.

(i)  $\text{10}$ cm

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=10cm}$

Hence substituting the values in the formula,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{10}}{\text{75}}$ radian

  $\text{=}\dfrac{\text{2}}{\text{15}}$radian

(ii) $\text{15}$ cm

And here, $\text{l=15cm}$  

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{15}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{1}}{\text{5}}$radian

(iii) $\text{21}$ cm

And here, $\text{l=21cm}$

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{21}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{7}}{\text{25}}$radian

Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ , $x$ lies in the third quadrant.

Ans: Here given that,  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\dfrac{\text{1}}{\text{cos x}}$

          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

          $\text{=-2}$

Now we know that,$\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\dfrac{\text{1}}{\text{4}}$

            $\text{=}\dfrac{\text{3}}{\text{4}}$

   $\text{sin x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$ 

                            $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}$ 

                            $\text{=-}\dfrac{\text{2}}{\sqrt{\text{3}}}$ 

 $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

        $\text{=}\dfrac{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

        $\text{=}\sqrt{\text{3}}$

$\text{cot x=}\dfrac{\text{1}}{\text{tan x}}$

       $\text{=}\dfrac{\text{1}}{\sqrt{\text{3}}}$

2. Find the values of other five trigonometric functions if $\text{sin  x=}\dfrac{\text{3}}{\text{5}}$  , $\text{x}$ lies in second quadrant.

Here given that,  $\text{sin x=}\dfrac{\text{3}}{\text{5}}$

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

          $=\dfrac{1}{\left( \dfrac{3}{5} \right)}$

           $=\dfrac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\dfrac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\dfrac{\text{9}}{\text{25}}$

         $\text{=}\dfrac{\text{16}}{\text{25}}$ 

   $\text{cos x= }\!\!\pm\!\!\text{ }\dfrac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\dfrac{\text{4}}{\text{5}}$

Therefore, $sec x=\dfrac{1}{\cos x}$ 

                          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$ 

                          $\text{=-}\dfrac{\text{5}}{\text{4}}$ 

 $\tan x=\dfrac{\sin  x}{\cos x}$ 

        $\text{=}\dfrac{\left( \dfrac{\text{3}}{\text{5}} \right)}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$

        $\text{=-}\dfrac{\text{3}}{\text{4}}$

$\cot x=\dfrac{1}{\tan x}$

        $\text{=-}\dfrac{\text{4}}{\text{3}}$

3. Find the values of other five trigonometric functions if   $\text{cot x=}\dfrac{\text{3}}{\text{4}}$ , $\text{x}$ lies in third quadrant.

Ans: Here given that,  $\cot x=\dfrac{3}{4}$

$\tan x=\dfrac{1}{\cot x}$

       $=\dfrac{1}{\left( \dfrac{3}{4} \right)}$

       $=\dfrac{4}{3}$

Now we know that , \[\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}\]

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\dfrac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \dfrac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{16}}{\text{9}}$

             $=\dfrac{25}{9}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{5}}{\text{3}}$

Therefore, $\cos  x=\dfrac{1}{\sec x}$ 

                         $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{3}} \right)}$ 

                          $\text{=-}\dfrac{\text{3}}{\text{5}}$ 

Now  , $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)$   

                                    \[\text{=}\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)\] 

 And 

          $\text{=-}\dfrac{\text{5}}{\text{4}}$

4. Find the values of other five trigonometric functions if $\text{sec  x=}\dfrac{\text{13}}{\text{5}}$  , $\text{x}$ lies in fourth quadrant.

Ans: Here given that,  $\sec x=\dfrac{13}{5}$

$\cos x=\dfrac{1}{\sec x}$

        $=\dfrac{1}{\left( \dfrac{13}{5} \right)}$

        $=\dfrac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  \[\text{sec x=}\dfrac{\text{13}}{\text{5}}\]  in the formula, we obtain,

\[\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \dfrac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}\]

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\dfrac{\text{169}}{\text{25}}\text{-1}$

         $\text{=}\dfrac{\text{144}}{\text{25}}$ 

   \[\text{tanx= }\!\!\pm\!\!\text{ }\dfrac{\text{12}}{\text{5}}\]

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\dfrac{\text{12}}{\text{5}}$

Therefore, \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\] 

                          $\text{=-}\dfrac{\text{5}}{\text{12}}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{5}} \right)$   

                                  $\text{=}\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$ 

            $\text{=-}\dfrac{\text{13}}{\text{12}}$

5. Find the values of other five trigonometric functions if  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$ , $\text{x}$ lies in second quadrant.

Ans: Here given that,  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$

       $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}$

       $\text{=-}\dfrac{\text{12}}{\text{5}}$

Substituting  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{25}}{\text{144}}$

             $=\dfrac{169}{144}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\dfrac{\text{1}}{\text{sec x}}$ 

                          $\text{=-}\dfrac{\text{12}}{\text{13}}$ 

 Hence we have, $\text{sin x=}\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$   

                                    $=\left( \dfrac{5}{13} \right)$ 

           $\text{=}\dfrac{\text{13}}{\text{5}}$

6. Find the value of the trigonometric function $\text{sin76}{{\text{5}}^{\text{o}}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi $ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

            $\text{=sin4}{{\text{5}}^{\text{o}}}$

            $\text{=}\dfrac{\text{1}}{\sqrt{\text{2}}}\text{.}$

7. Find the value of the trigonometric function $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)$  

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

           $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec3}{{\text{0}}^{\text{o}}}$

                               $=2$ 

8. Find the value of the trigonometric function  $\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

$\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\dfrac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

            $\text{=tan}\left( \text{6 }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

            \[\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

            $\text{=}\sqrt{\text{3}}$      

9. Find the value of the trigonometric function $\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

$\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}}\text{+2 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

                    $=\dfrac{\sqrt{3}}{2}$

10. Find the value of the trigonometric function $\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

Ans: We  know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

$\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

              $=1$

Exercise 3.3

1. Prove that $\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=-}\dfrac{\text{1}}{\text{2}}$

Ans: Substituting the values of  \[\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\] on left hand side,

$\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}$

                                  \[\text{=}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{4}}\text{-1}\]

                                  $=-\dfrac{1}{2}$

                                  $=$ R.H.S.

Hence proved.

2. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}$

Ans: Substituting the values of  $\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+cose}{{\text{c}}^{\text{2}}}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right){{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

           $\text{=2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{4}}\text{+}{{\left( \text{-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

          $\text{=}\dfrac{\text{1}}{\text{2}}\text{+}{{\left( \text{-2} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S  $=\dfrac{1}{2}+\dfrac{4}{4}$

           $=\dfrac{3}{2}$ 

          $=$ R.H.S.

3. Prove that $\text{co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=6}$

Ans: Substituting the values of  $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ on left hand side,

L.H.S.$\text{=co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

           $\text{=}{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{+cosec}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{+3}{{\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}$

         $\text{=3+cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{3}}$

we have, $\text{cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S $=3+2+1$ 

          $=1$ 

4. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=10}$

Substituting the values of  $\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{,sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left\{ \text{sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right\}}^{\text{2}}}\text{+2}{{\left( \dfrac{\text{1}}{\sqrt{\text{2}}} \right)}^{\text{2}}}\text{+2}{{\left( \text{2} \right)}^{\text{2}}}$

           $\text{=2}{{\left\{ \text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right\}}^{\text{2}}}\text{+2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}\text{+8}$

Since $\text{sin x}$ repeat its value after an interval of \[\text{2 }\!\!\pi\!\!\text{ }\] , 

we have, \[\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{=sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]  

 L.H.S  $=1+1+8$ 

            $=10$

            $=$ R.H.S.

5. Find the value of :

(i) $\text{sin7}{{\text{5}}^{\text{o}}}$

Ans: We have,

$\text{sin7}{{\text{5}}^{\text{o}}}\text{=sin(4}{{\text{5}}^{\text{o}}}\text{+3}{{\text{0}}^{\text{o}}}\text{)}$

          $\text{=sin4}{{\text{5}}^{\text{o}}}\text{cos3}{{\text{0}}^{\text{o}}}\text{+cos4}{{\text{5}}^{\text{o}}}\text{sin3}{{\text{0}}^{\text{o}}}$

Since we know that, $\text{sin}\left( \text{x+y} \right)\text{=sin x cos y+cos x sin y}$ 

$Sin 75^o = \dfrac{1}{\sqrt 2}\times \dfrac{\sqrt 3}{2} + \dfrac{1}{\sqrt 2}\times\dfrac{1}{2}$

$Sin 75^o = \dfrac{{\sqrt 3}+1}{2\sqrt 2}$

(ii) $\text{tan1}{{\text{5}}^{\text{o}}}$

\[\text{tan1}{{\text{5}}^{\text{o}}}\text{=tan}\left( \text{4}{{\text{5}}^{\text{o}}}\text{-3}{{\text{0}}^{\text{o}}} \right)\]

        $\text{=}\dfrac{\text{tan4}{{\text{5}}^{\text{o}}}\text{-tan3}{{\text{0}}^{\text{o}}}}{\text{1+tan4}{{\text{5}}^{\text{o}}}\text{tan3}{{\text{0}}^{\text{o}}}}$

Since we know, $\text{tan}\left( \text{x-y} \right)\text{=}\dfrac{\text{tan x-tan y}}{\text{1+tan x tan y}}$

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{1-}\dfrac{\text{1}}{\sqrt{\text{3}}}}{\text{1+1}\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}$

           $\text{=}\dfrac{\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}}}{\dfrac{\sqrt{\text{3}}\text{+1}}{\sqrt{\text{3}}}}$

         $\text{=}\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}\text{+1}}$

         $\text{=}\dfrac{{{\left( \sqrt{\text{3}}\text{-1} \right)}^{\text{2}}}}{\left( \sqrt{\text{3}}\text{+1} \right)\left( \sqrt{\text{3}}\text{-1} \right)}$ 

Further computing we have,

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{3+1-2}\sqrt{\text{3}}}{{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}}$ 

           \[\text{=}\dfrac{\text{4-2}\sqrt{\text{3}}}{\text{3-1}}\]

           \[\text{=2-}\sqrt{\text{3}}\]

6. Prove that $\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=sin}\left( \text{x+y} \right)$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos xcos y-sin xsin y}$ 

$\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x+}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right]$

$\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-}\left( \text{x+y} \right) \right]$

$\text{=sin}\left( \text{x+y} \right)$ 

L.H.S  $=$ R.H.S.

Hence  proved.

7. Prove that $\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}\text{=}{{\left( \dfrac{\text{1+tanx}}{\text{1-tanx}} \right)}^{\text{2}}}$

Ans: We  know that ,$\text{tan}\left( \text{A+B} \right)\text{=}\dfrac{\text{tan A+tan B}}{\text{1-tan Atan B}}$

and $\text{tan}\left( \text{A-B} \right)\text{=}\dfrac{\text{tan A-tan B}}{\text{1+tan Atan B}}$

L.H.S.$\text{=}\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}$

Using the above formula,

          $\text{L}\text{.H}\text{.S=}\dfrac{\left( \dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+tanx}}{\text{1-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}} \right)}{\dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-tanx}}{\text{1+tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}}}$

                    $\text{=}\dfrac{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}{\left( \dfrac{\text{1-tan x}}{\text{1+tan x}} \right)}$      [ substituting $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=1}$ ]

                    $\text{=}{{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}^{\text{2}}}$

                    $=$ R.H.S.

8.  Prove that  $\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

Ans: Observe that $\text{cos x}$ repeats same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ 

and $\text{sin x}$ repeats same value after an interval  $\text{2 }\!\!\pi\!\!\text{ }$.

L.H.S.$\text{=}\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}$

           $\text{=}\dfrac{\left[ \text{-cos x} \right]\left[ \text{cos x} \right]}{\left( \text{sin x} \right)\left( \text{-sin x} \right)}$

           $\text{=}\dfrac{\text{-co}{{\text{s}}^{\text{2}}}\text{x}}{\text{-si}{{\text{n}}^{\text{2}}}\text{x}}$

           $\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

9. Prove that,

$\text{Cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)\text{Cos}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right)\left[ \text{cot}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{-x} \right)\text{+cot}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right) \right]\text{=1}$

Ans: We know that $\text{cot x}$  repeats same value after an interval $2\pi $ .

L.H.S.$=Cos\left( \dfrac{3\pi }{2}+x \right)Cos\left( 2\pi +x \right)\left[ cot\left( \dfrac{3\pi }{2}-x \right)+cot\left( 2\pi +x \right) \right]$

           $\text{=sin x cos x}\left[ \text{tan x+cot x} \right]$

Substituting $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ and

$\text{cot x=}\dfrac{\text{cos x}}{\text{sin x}}$ ,

$\text{L}\text{.H}\text{.S=sin xcos x}\left( \dfrac{\text{sin x}}{\text{cos x}}\text{+}\dfrac{\text{cos x}}{\text{sin x}} \right)$

         $\text{=}\left( \text{sin x cos x} \right)\left[ \dfrac{\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x}}{\text{sin x cos x}} \right]$

         $\text{=1}$ 

        $\text{=}$ R.H.S.

10. Prove that $\text{sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x=cos x}$

Ans: We know that , $\text{cos}\left( \text{x-y} \right)\text{=cosxcosy+sinxsiny}$ 

L.H.S.$\text{=sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x}$

           $\text{=cos}\left[ \left( \text{n+1} \right)\text{x-}\left( \text{n+2} \right)\text{x} \right]$ 

           $\text{=cos}\left( \text{-x} \right)$ 

           $\text{=cosx}$ 

           $=$  R.H.S.

11. Prove that $\text{cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{=-}\sqrt{\text{2}}\text{sinx}$

Ans: We  know that , $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)$

               $\text{=-2sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{+}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}\text{.sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}$

               $\text{=-2sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{sin x}$ 

Since $\text{sin x}$ repeats the same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

therefore, 

$\text{L}\text{.H}\text{.S=-2sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{sin x}$ 

           $\text{=-2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{2}}}\text{ }\!\!\times\!\!\text{ sinx}$

           $\text{=-}\sqrt{\text{2}}\text{sin x}$

           $\text{=}$ R.H.S.

12. Prove that $\text{si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4x=sin 2x sin 10x}$

Ans: We know that,$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4xa}$

               $\text{=}\left( \text{sin 6x+sin 4x} \right)\left( \text{sin 6x-sin 4x} \right)$

               $\text{=}\left[ \text{2sin}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]\left[ \text{2cos}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]$

               $\text{=}\left( \text{2sin 5x cos x} \right)\left( \text{2cos 5x sin x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ ,

$\text{L}\text{.H}\text{.S=}\left( \text{2sin 5x cos 5x} \right)\left( \text{2sin x cos x} \right)$ 

          $\text{=sin 10x sin 2x}$

          $\text{=}$ R.H.S.

13. Prove that $\text{co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x=sin 4x sin 8x}$

Ans: We  know that,

$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x}$

            $\text{=}\left( \text{cos 2x+cos 6x} \right)\left( \text{cos 2x-6x} \right)$

            $\text{=}\left[ \text{2cos}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\left[ \text{-2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]$

Further computing, we have,

$\text{L}\text{.H}\text{.S=}\left[ \text{2cos 4x cos}\left( \text{-2x} \right) \right]\left[ \text{-2sin 4xsin}\left( \text{-2x} \right) \right]$

         $\text{=}\left[ \text{2cos  4x cos 2x} \right]\left[ \text{-2sin 4x}\left( \text{-sin 2x} \right) \right]$

         $\text{=}\left( \text{2sin 4x cos 4x} \right)\left( \text{2sin 2xcos 2x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ 

Therefore we have,                

$\text{L}\text{.H}\text{.S=sin 8x sin 4x}$

         $\text{=}$  R.H.S.

.Hence proved.

14. Prove that $\text{sin 2x+2sin 4x+sin6=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

Ans: We know that, \[\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

L.H.S.$\text{=sin 2x+2sin 4x+sin 6x}$

.           $\text{=}\left[ \text{sin 2x+sin 6x} \right]\text{+2sin 4x}$

           $\text{=}\left[ \text{2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\text{+2sin4x}$

           $\text{=2sin 4xcos}\left( \text{-2x} \right)\text{+2sin 4x}$

We have, $\text{L}\text{.H}\text{.S=2sin 4x cos 2x+2sin 4x}$ 

                         $\text{=2sin 4x}\left( \text{cos 2x+1} \right)$ 

Now we know that, $\text{cos 2x+1=2co}{{\text{s}}^{\text{2}}}\text{x}$ 

$\text{L}\text{.H}\text{.S=2sin 4x}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)$ 

         $\text{=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

 = R.H.S.

15. Prove that $\text{cot 4x}\left( \text{sin 5x+sin 3x} \right)\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

Ans: We know that, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

L.H.S.$\text{=cot 4x}\left( \text{sin 5x+sin 3x} \right)$

          $\text{=}\dfrac{\text{cot 4x}}{\text{sin 4x}}\left[ \text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right) \right]$

          $\text{=}\left( \dfrac{\text{cos 4x}}{\text{sin 4x}} \right)\left[ \text{2sin 4x cos x} \right]$

          $\text{=2cos 4x cos x}$

Now also ,we know that, $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 R.H.S.$\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{5x-3x}}{\text{2}} \right) \right]$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos 4x sin x} \right]$

            $\text{=2cos 4x cos x}$

Therefore , we can conclude that,

L.H.S.=R.H.S.

16. Prove that $\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$

$\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=}\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}$

            \[\text{=}\dfrac{\text{-2sin}\left( \dfrac{\text{9x+5x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{9x-5x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{17x+3x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{17x-3x}}{\text{2}} \right)}\]                    (following the formula)

           

            $\text{=}\dfrac{\text{-2sin 7x}\text{.sin 2x}}{\text{2cos 10x}\text{.sin 7x}}$

            $\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$ 

17. Prove that:$\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}\text{=tan 4x}$

We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

Now , L.H.S.$\text{=}\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$                  (using the formula)

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$

                      $\text{=}\dfrac{\text{2sin 4x cos x}}{\text{2cos 4x cos x}}$

$\text{L}\text{.H}\text{.S=tan 4x}$ 

18. Prove that \[\dfrac{\text{sin x-sin y}}{\text{cos x+cos y}}\text{=tan}\dfrac{\text{x-y}}{\text{2}}\]

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

.$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

  L.H.S.\[\text{=}\dfrac{\text{sin x-sin y}}{\text{cosx+cosy}}\]

               $\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

               $\text{=}\dfrac{\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

              $\text{=tan}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Therefore $\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}$ 

19. Prove that $\dfrac{\text{sin x+sin 3x}}{\text{cos x+cos 3x}}\text{=tan 2x}$

Ans: We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{,}$

.\[\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

Now , L.H.S.$\text{=}\dfrac{\text{sinx+sin3x}}{\text{cos x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}$                   (using the formula)

                      $\text{=}\dfrac{\text{sin 2x}}{\text{cos 2x}}$

                      $\text{=tan 2x}$

Therefore  L.H.S$=$ R.H.S.

20. Prove that $\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2sin x}$

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And \[\text{co}{{\text{s}}^{\text{2}}}\text{A-si}{{\text{n}}^{\text{2}}}\text{A=cos 2A}\]

 L.H.S.$\text{=}\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}$

       \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{-cos2x}}\]

            $\text{=}\dfrac{\text{2cos2xsin}\left( \text{-x} \right)}{\text{-cos 2x}}$

            $\text{=-2 }\!\!\times\!\!\text{ }\left( \text{-sinx} \right)$

Therefore , we have,

$\text{L}\text{.H}\text{.S=2sin x}$

        $=$ R.H.S.

21. Prove that $\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}\text{=cot 3x}$

And, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Now, L.H.S.$\text{=}\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}$

                     \[\text{=}\dfrac{\left( \text{cos 4x+cos 2x} \right)\text{+cos 3x}}{\left( \text{sin4x+sin2x} \right)\text{+sin 3x}}\]

                     \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+cos3x}}{\text{2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+sin 3x}}\]   (using the formulas)

                      \[\text{=}\dfrac{\text{2cos 3x cos x+cos 3x}}{\text{2sin 3x cos x+sin 3x}}\]

Further computing, we obtain,

L.H.S$\text{=}\dfrac{\text{cos 3x}\left( \text{2cos x+1} \right)}{\text{sin 3x}\left( \text{2cos x+1} \right)}$ 

          \[\text{=cot 3x}\] 

22. Prove that \[\text{cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1}\]

We know that, \[\text{cot}\left( \text{A+B} \right)\text{=}\dfrac{\text{cotAcotB-1}}{\text{cot A+cot B}}\]

Now , L.H.S.$\text{=cot xcot 2x-cot 2x cot 3x-cot 3x cot x}$

                      \[\text{=cot x cot 2x-cot 3x}\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-cot}\left( \text{2x+x} \right)\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-}\left[ \dfrac{\text{cot 2x cot x-1}}{\text{cot x+cot 2x}} \right]\left( \text{cot 2x+cot x} \right)\]

Further computing we obtain,

$\text{L}\text{.H}\text{.S=cot x cot 2x-}\left( \text{cot 2x cot x-1} \right)$ 

         \[\text{=1}\]

         $\text{=}$ R.H.S.

23. Prove that $\text{tan 4x=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$

Ans: We  know that $\text{tan 2A=}\dfrac{\text{2tan A}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{A}}$

L.H.S.$\text{=tan 4x}$

           $\text{=tan2}\left( \text{2x} \right)$

           \[\text{=}\dfrac{\text{2tan 2x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\left( \text{2x} \right)}\][using the formula]

           $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \text{1-}\dfrac{\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$

Further  computing, we obtain,

L.H.S $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \dfrac{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$$$$$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1+ta}{{\text{n}}^{\text{4}}}\text{x-2ta}{{\text{n}}^{\text{2}}}\text{x-4ta}{{\text{n}}^{\text{2}}}\text{x}}$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$ 

24. Prove that $\text{cos 4x=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$

Ans: We know that, $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$ 

And $\text{sin 2x=2sin x cos x}$ 

L.H.S.\[\text{=cos 4x}\]

           $\text{=cos 2}\left( \text{2x} \right)$

           $\text{=1-2si}{{\text{n}}^{\text{2}}}\text{2x}$

           $\text{=1-2}{{\left( \text{2sin x cos x} \right)}^{\text{2}}}$

Further computing we get,

L.H.S$\text{=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$ 

          $=$R.H.S.

25. Prove that $\text{cos 6x=32xco}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

Ans: We know that, $\text{cos 3A=4co}{{\text{s}}^{\text{3}}}\text{A-3cosA}$

and  $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$

L.H.S.$\text{=cos 6x}$

           $\text{=cos 3}\left( \text{2x} \right)$

           \[\text{=4co}{{\text{s}}^{\text{3}}}\text{2x-3cos 2x}\]

           \[\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right) \right]\]

L.H.S$\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}{{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{+3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ \text{8co}{{\text{s}}^{\text{6}}}\text{x-1-12co}{{\text{s}}^{\text{4}}}\text{x+6co}{{\text{s}}^{\text{2}}}\text{x} \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=32co}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

L.H.S $=$ R.H.S.

Exercise 3.4

1. Find the principal and general solutions of the \[\text{tan x=}\sqrt{\text{3}}\].

Ans: Here given that,

\[\text{tan x=}\sqrt{\text{3}}\]

We know that $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

and $\text{tan}\left( \dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

Therefore, the principal solutions are\[\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\] and \[\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\] .

Now, \[\text{tan x=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where \[\text{n}\in \text{Z}\]

Therefore, the general solution is \[\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\], where $\text{n}\in \text{Z}$ .

2. Find the principal and general solutions of the equation $\text{secx=2}$

Ans: Here it is given that,

$\text{sec x=2}$

Now we know that

$\text{sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=2}$ and 

$\text{sec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=sec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

          $\text{=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

          $\text{=2}$ 

Therefore, the principal solutions are$\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ and \[\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\].

Now, $\text{sec x=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

and we know , 

$\sec x=\dfrac{1}{\cos  x}$

\[\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

$\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $n\in Z$ .

3. Find the principal and general solutions of the equation $\text{cot x=-}\sqrt{\text{3}}$

\[\text{cot x=-}\sqrt{\text{3}}\]

Now we know that $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=}\sqrt{\text{3}}$

\[\text{cot}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                 \[\text{=-}\sqrt{\text{3}}\]

and  $\text{cot}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

                          \[\text{=-}\sqrt{\text{3}}\]

$\text{cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$ 

and $\text{cot}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$

Therefore, the principal solutions are $\text{x=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ and $\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}$.

.Now, $\text{cot x=cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\]

$\text{tan x=tan}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$ .

4. Find the general solution of $\text{cosec x=-2}$

$\text{cosec x=-2}$

$\text{cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=2}$

$\text{cosec}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{a}$ 

                     $\text{=-2}$

and \[\text{cosec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                                        \[\text{=-2}\]

therefore we have,

\[\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}\]and $\text{cosec}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}$

Hence , the principal solutions are$\text{x=}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\,$ and $\text{ }\dfrac{11\pi }{6}$.

Now, $\text{cosec x=cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know, \[\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}\]

$\text{sin x=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  

,where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{ }$ ,where $\text{n}\in \text{Z}$.

5. Find the general solution of the equation $\text{cos 4x=cos 2x}$

Ans: Here it is given that, $\text{cos 4x=cos 2x}$

$\text{cos 4x-cos 2x=0}$

Now we know that, $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$ 

$\text{-2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{=0}$

                            $\text{sin 3x sin x=0}$

Hence we have, $\text{sin 3x=0}\,\,$

Or, $\text{ sin x=0}$

Therefore, $\text{3x=n }\!\!\pi\!\!\text{ }$

 or    $\text{x=n }\!\!\pi\!\!\text{ }$    ,where $\text{ n}\in \text{Z}$

 therefore, \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\]     

 or  $\text{x=n }\!\!\pi\!\!\text{ }$  ,where $\text{ n}\in \text{Z}$.

6. Find the general solution of the equation $\text{cos 3x+cos x-cos 2x=0}$.

$\text{cos 3x+cos x-cos 2x=0}$

Now we know that, $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore  $\text{cos 3x+cos x-cos 2x=0}$ implies

$\text{2cos}\left( \dfrac{\text{3x+x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-x}}{\text{2}} \right)\text{-cos 2x=0}$

                     $\text{2cos 2x cos x-cos 2x=0}$

                        $\text{cos 2x}\left( \text{2cos x-1} \right)\text{=0}$

Hence we have, 

 Either $\text{cos 2x=0}$

Or $\text{cos x=}\dfrac{\text{1}}{\text{2}}$ 

Which in turn implies that,

Either $\text{2x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,$

Or,  $\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$    , where  $\text{n}\in \text{Z}$

Either  $\text{x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4}\,\,$

Or,  $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  ,where  \[\text{n}\in \text{Z}\].

7. Find the general solution of the equation \[\text{sin 2x+cos x=0}\] .

\[\text{sin 2x cos x=0}\]

\[\text{2sin x cos x+cos x=0}\]

     \[\text{cos x(2sin x+1)= }\!\!~\!\!\text{ 0}\]

Either $\text{cos x=0}$ 

Or, $\text{sin x=-}\dfrac{\text{1}}{\text{2}}$  

Hence we have,

 \[\text{x= }\!\!~\!\!\text{ (2n+1)}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\]   , where $\text{n}\in \text{Z}$ .

Either 

Or,  \[\text{sin x=-}\dfrac{\text{1}}{\text{2}}\]

             $\text{=-sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ 

             $\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$ 

             $\text{=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ 

Which implies

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  , where  $\text{n}\in \text{Z}$

Therefore, the general solution is \[\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,\] or  \[\text{n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,n}\in \text{Z}\].

8. Find the general solution of the equation \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Ans: Here given that , \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Now we know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$ 

      \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]  implies

           \[\text{1+ta}{{\text{n}}^{\text{2}}}\text{2x=1-tan 2x}\]

   \[\text{ta}{{\text{n}}^{\text{2}}}\text{2x+tan 2x=0}\]

 \[\text{tan 2x(tan 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence  either $\text{tan 2x=0}$ 

Or, $\text{tan 2x=-1}$ 

Which implies  either  \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\]  , where $\text{n}\in \text{Z}$ ,

               $\text{=-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

               $\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

                $\text{=tan}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

Which in turn implies that,  

$\text{2x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,}$ where  $\text{n}\in \text{Z}$

i.e,  $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,}$ where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\,\,$ or  $\,\,\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,n}\in \text{Z}$.

9. Find the general solution of the equation \[\text{sin x+sin 3x+sin 5x=0}\]

Here given that ,\[\text{sin x+sin 3x+sin 5x=0}\]

Now we know that,  $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore ,

                             \[\text{sin x+sin 3x+sin 5x=0}\]

                          \[\left( \text{sin x+sin 3x} \right)\text{+sin 5x=0}\] 

     $\left[ \text{2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right) \right]\text{+sin 3x=0}\,$

                       \[\text{2sin 3x cos (-2x)+sin 3x= }\!\!~\!\!\text{ 0}\]

Simplifying we get,

\[\text{2sin 3xcos 2x+sin 3x=0}\]

     \[\text{sin 3x(2cos 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence either $\text{sin 3x=0}$ 

Or, $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$ 

Which implies  $\text{3x=n }\!\!\pi\!\!\text{ }$ , where $\text{n}\in \text{Z}$ 

Or,   $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$

                  $\text{=-cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

                  $\text{=cos}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

                     $\text{=cos}\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$ 

i.e., either $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}$   , where $\text{n}\in \text{Z}$ 

or,  $\text{2x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$   ,where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\,$ or $\text{n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{,n}\in \text{Z}$.

Miscellaneous Exercise

1. Prove that: $\text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}\text{=0}$

Ans: We know that $\text{cos x+cos y=2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Now L.H.S.$\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)$       (using the formula)

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{- }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

Simplifying, 

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}} \right]$

         $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\left( \dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right]$

          $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}} \right]$

Substituting $\text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{=0}$ , we get,

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{ }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}}$ 

           $\text{=0}$

           $=\text{R}\text{.H}\text{.S}$

2. Prove that: $\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x=0}$

We know that, $\text{sin x+sin y=2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

And  $\text{cos x-cos y=-2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

L.H.S.$\text{=}\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x}$

           $\text{=sin 3x sin x+si}{{\text{n}}^{\text{2}}}\text{x+cos 3x cos x-co}{{\text{s}}^{\text{2}}}\text{x}$    (using the formula)

           $\text{=cos 3x cos x+sin 3x sin x-}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x-si}{{\text{n}}^{\text{2}}}\text{x} \right)$

  Simplifying  we get,       

$\text{L}\text{.H}\text{.S=cos}\left( \text{3x-x} \right)\text{-cos 2x}\,$

         $\text{=cos 2x-cos 2x}$

         $\text{=0}$

         $=\text{R}\text{.H}\text{.S}\text{.}$

3. Prove that: ${{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4co}{{\text{s}}^{\text{2}}}\dfrac{\text{x+y}}{\text{2}}$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos x cos y-sin xsin y}$

and  L.H.S$\text{=}{{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

                  \begin{align} & \text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y+2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y} \\ &  \\ \end{align}

                 $\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{+2}\left( \text{cos x cos y-sin x sin y} \right)$

Simplifying and using the formula,

L.H.S$\text{=1+1+2cos}\left( \text{x+y} \right)$ 

         $\text{=2}\left[ \text{1+cos}\left( \text{x+y} \right) \right]$ 

         $\text{=2}\left[ \text{1+2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1} \right]$ 

 [since $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1=cos}\left( \text{x+y} \right)$ ]

          $\text{=4co}{{\text{s}}^{\text{2}}}\left( \text{x+y} \right)$ 

Therefore  L.H.S$=$ R.H.S

4. Prove that:   ${{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4si}{{\text{n}}^{\text{2}}}\dfrac{\text{x-y}}{\text{2}}$ 

Ans: We know that, $\text{cos}\left( \text{x-y} \right)\text{=cos x cos y+sin x sin y}$ 

L.H.S.$\text{=}{{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

           $\text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y-2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y}$

           \[\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{-2}\left[ \text{cos x cos y+sin x sin y} \right]\]

Simplifying and using the formula  we get,

         L.H.S $\text{=1+1-2}\left[ \text{cos}\left( \text{x-y} \right) \right]\,\,$

                   $\text{=2}\left[ \text{1-cos}\left( \text{x-y} \right) \right]$

                   $\text{=2}\left[ \text{1-}\left\{ \text{1-2si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right) \right\} \right]\,$ 

[since  $\text{1-2si}{{\text{n}}^{\text{2}}}\dfrac{\left( \text{x-y} \right)}{\text{2}}\text{=cos}\left( \text{x-y} \right)$ ]

                   $\text{=4si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

5. Prove that: $\text{sin x+sin 3x+sin 5x+sin 7x=4cos xcos 2xsin 4x}$

Ans: We  know  that $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =sin x+sin 3x+sin 5x+sin 7x}$

            \[\text{=}\left( \text{sin x+sin 5x} \right)\text{+}\left( \text{sin 3x+sin 7x} \right)\]      

Using the formula and simplifying,

          $\text{=2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right)\text{+2sin}\left( \dfrac{\text{3x+7x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-7x}}{\text{2}} \right)$       

          \[\text{=2cos 2x}\left[ \text{sin 3x+sin 5x} \right]\]

          \[\text{=2cos 2x}\left[ \text{2sin}\left( \dfrac{\text{3x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{3x-5x}}{\text{2}} \right) \right]\] 

          \[\text{=2cos 2x}\left[ \text{2sin 4x}\text{.cos}\left( \text{-x} \right) \right]\] 

\[\text{L}\text{.H}\text{.S=4cos 2x sin 4x cos x}\] 

           \[=\text{R}\text{.H}\text{.S}\]

6. Prove that: $\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin 9x+sin 3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos 9x+cos 3x} \right)}\text{=tan 6x}$

Ans: We  known that,

$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =}\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin9x+sin3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos9x+cos3x} \right)}$

          $\text{=}\dfrac{\left[ \text{2sin}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2sin}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}{\left[ \text{2cos}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2cos}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}$

          $\text{=}\dfrac{\left[ \text{2sin 6x}\text{.cos x} \right]\text{+}\left[ \text{2sin 6x}\text{.cos 3x} \right]}{\left[ \text{2cos 6x}\text{.cos x} \right]\text{+}\left[ \text{2cos 6x}\text{.cos 6x} \right]}$

          $\text{=}\dfrac{\text{2sin 6x}\left[ \text{cos x+cos 3x} \right]}{\text{2cos 6x}\left[ \text{cos x+cos 3x} \right]}$

           $\text{=tan 6x}$

Therefore L.H.S$=$ R.H.S

7. Prove that: 

$\text{sin 3x+sin 2x-sin x=4sin xcos}\dfrac{\text{x}}{\text{2}}\text{cos}\dfrac{\text{3x}}{\text{2}}$

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)$ 

$\text{L}\text{.H}\text{.S}\text{.=sin3x+sin2x-sinx}$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{2x+x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-x}}{\text{2}} \right) \right]\,$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

Since we know that, $\text{sin 2x=2sin xcos x}$ 

$\text{L}\text{.H}\text{.S=2sin}\dfrac{\text{3x}}{\text{2}}\text{.cos}\dfrac{\text{3x}}{\text{2}}\text{+2cos}\dfrac{\text{3x}}{\text{2}}\text{sin}\dfrac{\text{x}}{\text{2}}\,\,\,\,\,\,$

         $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{sin}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

         \[\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{2sin}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\}\text{cos}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{-}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\} \right]\]

          $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{.2sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)$

Therefore 

 $\text{L}\text{.H}\text{.S=4sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x}}{\text{2}} \right)$

          $\text{=R}\text{.H}\text{.S}$ 

8.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{tanx=-}\dfrac{\text{4}}{\text{3}}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ is in 2nd quadrant.

$\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{x }\!\!\pi\!\!\text{ }$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}<\dfrac{\text{x}}{\text{2}}<\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, \[\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,\] and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{tan x=-}\dfrac{\text{4}}{\text{3}}$

We know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

         $\text{=1+}{{\left( \text{-}\dfrac{\text{4}}{\text{3}} \right)}^{\text{2}}}$ 

         \[\text{=}\dfrac{\text{25}}{\text{9}}\]

As \[\text{x}\] is in 2nd quadrant, $\text{sec x}$ is negative.

Therefore , $\text{secx=-}\dfrac{\text{5}}{\text{3}}$ 

Then $\text{cos x=-}\dfrac{\text{3}}{\text{5}}$ 

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cos x+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{5}}$ 

Hence \[\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}\] 

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\sqrt{\text{5}}}$ 

$\text{tan}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{sin}\dfrac{\text{x}}{\text{2}}}{\text{cos}\dfrac{\text{x}}{\text{2}}}$ 

          $=2$ 

Thus, the respective values of$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are $\,\dfrac{2\sqrt{5}}{5},\dfrac{\sqrt{5}}{5},\,\,2$ .

9.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\cos x\text{=-}\dfrac{1}{\text{3}}$ , $\text{x}$ lies in 3rd quadrant.   

Ans: Here, $\text{x}$ is in 3rd quadrant.

$\text{  }\!\!\pi\!\!\text{ x}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{}\dfrac{\text{x}}{\text{2}}\text{}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 2nd quadrant.

Therefore, $\text{cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are negative and $\text{sin}\dfrac{\text{x}}{\text{2}}$ is positive.

 Given that $\text{cos x=-}\dfrac{1}{\text{3}}$

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cosx+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{3}}$ 

Hence $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{2}}{\text{3}}}$ 

          $\text{=-}\sqrt{\text{2}}$ 

Thus, the respective values of $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are  $\,\sqrt{\dfrac{\text{2}}{\text{3}}}\text{,-}\dfrac{\text{1}}{\sqrt{\text{3}}}\text{,}\,\text{-}\,\sqrt{\text{2}}$.

10.Find$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{sin x=}\dfrac{1}{4}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ lies in 2nd quadrant.

hence  $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{sin x=}\dfrac{\text{1}}{\text{4}}$

Therefore substituting $\text{sin x=}\dfrac{\text{1}}{\text{4}}$ and computing ,

$\text{cos x=-}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

since $\text{x}$ lies in 2nd quadrant,  $\text{cos x}$ is negative.

Now we know that, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1-cos x}$ 

Computing we get, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1+}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

Hence $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ 

Therefore substituting $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ and computing ,

$\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}$ 

          $\text{=}\dfrac{\sqrt{\text{4+}\sqrt{\text{15}}}}{\sqrt{\text{4-}\sqrt{\text{15}}}}$ 

       \[\text{=4+}\sqrt{\text{15}}\] 

are  $\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}\text{,}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}\text{,4+}\sqrt{\text{15}}$ .

NCERT Solutions for CBSE Class 11 Maths Chapter 3 Exercise 3.2 Trigonometric Functions

Ncert solutions for class 11 maths chapter 3 – free pdf download.

The Mathematical jargon in Trigonometry Class 11 is undoubtedly a complex affair for most students to grasp. But at the same time, it is an indispensable topic in 11 th standard syllabus. Stronghold on this topic constitutes a very basic understanding of geometric calculations and is crucial for solving multiple numerical problems in other subjects besides Mathematics.

NCERT Solutions for Class 11 Maths Chapter 3 are formulated by experienced and highly learned tutors focusing specifically on the problematic areas students tend to encounter in this chapter. Numerical sums based on the application of trigonometric formulas, graphs of trigonometric functions, conversion of angles from radian to degree and vice versa, and general solutions have been explained in great detail and elaborate steps.

Additionally, these study guides come with topic-wise exercises and probable questions as well, because increased practice will only help students improve their problem-solving speed and accuracy. In order to avail these suggestions offline, you can click on NCERT Solutions for Class 11 Maths Chapter 3 PDF Download option on Vedantu’s website.

NCERT Solutions for Class 11 Maths Chapter 3 Subtopics

Before wandering into the conceptual details of Trigonometric Functions Class 11, you can have a look at the summary of all topics discussed in this chapter.

Section 1: Introduction

In this part, students are introduced to fundamental trigonometric functions and their application. You will be required to perform calculations of distances based on these ratios.

Section 2: Angles

Class 11 Maths Trigonometry comes with four subsections under this topic. Students will learn about measurement of angles in different units, radians and degrees, and relations between them. Solutions to numerical sums involving the conversion of angle measurements from one form to another have been discussed in detail in NCERT Solutions for Class 11 Maths Chapter 3.

Section 3: Trigonometric Functions

This section deals with two topics. First one teaches the signs and symbols of generalised trigonometric functions in all four quadrants of a graph. In the next subsection, students will be acquainted with the domain and range of the different functions, that is, how the value of a function increases or decreases for an increasing angle value. You can find tables and detailed explanations on the working of these concepts in Trigonometric Functions Class 11 NCERT Solutions.

Section 4: Trigonometric Functions of Sum and Difference of Two Angles

The concluding section of Chapter 3 Maths Class 11 PDF focuses on the derivation of formulas and expressions based on trigonometric functions of the sums and differences between two angles. These formulas have been explained with the help of examples. Students must have a clear understanding of these as they make up an important part of geometric evaluations and are applied in a wide range of calculations.

Important Points Covered in Class 11 Maths Chapter 3 - Trigonometric Functions

The term “Angle” refers to a measurement of rotation of a ray about its fixed end point. The initial position of the ray before starting the rotation is called the initial side of the angle and the final position of the ray after rotation is called the terminal side of the angle.

In the Sexagesimal system, the angles are measured in degrees. 

1° = 60′ reads as 1 degree is equal to 60 minutes. 

1′ = 60″ reads as 1 minute is equal to 60 seconds.

In the Circular system, the angles are measured in radians.

1 Radian = 180/π degree and 

1 Degree = π/180 radians

Functions of negative angles (A)

sin (-A) = - sinA,

cosec (-A) = - cosec A,

tan (-A) = - tan A,

cos (-A) = cos A,

cot (-A) = - cot A,

sec (-A) = sec A.

Note: The trigonometric functions cosine and secant are even functions. The remaining all trigonometric functions are odd functions.

Trigonometric formulas of compound angles:

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

2 sin A cos B = sin (A + B) + sin (A – B)

2 cos A sin B = sin (A + B) – sin (A – B)

2 cos A cos B = cos (A + B) + cos (A – B)

2 sin A sin B = cos (A – B) – cos (A + B)

Trigonometric equations are the equations which include trigonometric functions of unknown angles.

Solutions of a trigonometric equation are all the possible values of the unknown angle that satisfies the given equation. These solutions (possible values) can be an infinite number in some cases.

The general solution of a trigonometric equation is defined as the solution that involves an integer ‘n’ which gives all solutions of the given trigonometric equation.

The principal solution of a trigonometric equation is defined as the solutions which lie in the interval 0 ≤ x ≤ 2π.

We Cover All The Exercises In The Chapter Given Below

Key features of ncert solutions for class 11 maths chapter 3 - trigonometric functions.

Key features of NCERT Solutions for Class 11 Maths Chapter 3 are as follows:

Pythagorean Identities will be used to solve many trigonometric problems in which one trigonometric ratio is given and the other ratios must be calculated.

We will learn the relationship between the ratios of the sides of a right-angled triangle and their angles.

We will learn about trigonometric ratios such as sine, cosine, tangent, cotangent, secant, and cosecant. 

The solutions deal with angle measurement and angle problems. 

We use fundamental ratios to construct other key trigonometric functions: cotangent, secant, and cosecant. These functions serve as the foundation for all of the essential trigonometry topics.

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FAQs on NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

1. What are the basic trigonometric functions?

In Mathematics, trigonometric functions are fundamental functions used to denote relation of the angles of a right-angled triangle to the length of its sides.

There are six trigonometric ratios in total, the basic three being sine, cosine, and tangent. The other three are described in relation to previously-discussed basic functions and are called cosecant, secant and cotangent.

In a given right-angled triangle, values of these ratios are calculated for a specific angle θ in terms of its sides in the following manner.

sin θ = opposite side / hypotenuse

cos θ = adjacent side / hypotenuse

tan θ = sin θ / cos θ = opposite site / adjacent side

cosec θ = 1 / sin θ

sec θ = 1 / cos θ

cot θ = cos θ / sin θ = 1 / tan θ

2. How are Radians and Degrees related?

A Radian (denoted by rad or c) is a standard unit of measuring angles whose value is based on the relationship between the length of an arc and radius of a circle.

rad = arc length / radius of circle

In a lot of cases, you might be required to convert given angle values in Radian into Degree. For that, the relation between these two units can be derived as follows.  

For a full circle with radius r and angle 360 o , the arc length is equal to circumference C.

Therefore, arc length = C = 2πr.

From the first relation, 360 o in radians will be

rad = 2πr / r = 2π

Therefore, 2π rad = 360 o ,

that is, 1 rad = 180 o / π.

This brings us to the final relationship that is, degree = radian x 180 / π.

3. What can you learn from Trigonometry Class 11 NCERT Solutions PDF?

Chapter 3 Maths Class 11 covers the vast and complex topic of trigonometric functions and their applications. This chapter comes with a total of four subsections dealing with concepts like measuring angles in degrees and radians and their interconversion, sine and cosine formulas in terms of variable angles x and y, finding solutions of trigonometric values, and so on.

NCERT Solutions for Class 11 Maths Chapter 3 comes with comprehensive answers to questions from all four exercises given in textbooks. Solutions to numerical sums have been explained in a step-by-step manner. You will learn about the applications of trigonometric equations in the vast field of science and be able to recognise these functions in Physics and Chemistry.

4. How do you solve trigonometric functions in Class 11 Maths?

Trigonometry can be a tricky topic. However, the students who have a good grasp of the basics of trigonometry seem to solve the problems in a better way. The first thing to remember while solving the trigonometric functions is to know the formulae and their applications. Vedantu offers a complete guide and the PDF of the solutions of the exercise given in the NCERT textbook. The solutions are provided free of cost in a step-by-step manner and are easy to comprehend.

5. Is Chapter 3 of Class 11 Maths hard?

A lot of students may find Trigonometry difficult. However, if the concepts are clear and they have a good hold on the basics of Trigonometry, then it seems easy. As trigonometry is all about the relation between the angles and sides of the triangle, students need to know the basic formulae and their applications to solve the problems. Vedantu offers solutions that are verified by the subject-matter-expert and are also solved comprehensively. 

6. What are the important topics in Chapter 3 of Class 11 Maths?

In Class 11 Mathematics, Chapter 3 is all about trigonometry. This chapter includes all about the relation between the angles and the sides of the triangles and their applications. Whenever there is a need to solve the trigonometric functions, a student must have a good grasp of the identities and the formula. The value table of all the trigonometric functions for different angles must be remembered for solving problems. Vedantu offers the solution to all the exercises provided in the NCERT Textbook in a step-by-step manner. 

7. What is Trigonometry for Class 11 Students?

Trigonometry is a section of Mathematics, which provides an understanding of the relation between the angles and the sides of the triangle. This chapter can be tricky for most students. But, the students can ace their exams by practising numerous questions and knowing the formulae and their applications. Vedantu offers the best learning guide and the PDF of the solutions that are verified by the subject-matter-experts. Students can either refer to it online or by downloading the PDF for free to refer to it offline. 

8. How can Trigonometry of Class 11 be made easy to study?

Trigonometry seems difficult for most of the students. Few of the ways to make it easy is to always choose the side that is complex, denote or represent all the trigonometric functions into sine and cosine, focus on the formulae and the signs, and grasp the basics of cancelling the terms, rationalizing, and expanding. Vedantu offers an explanation based solution that is easy to follow and score. 

9. What are the applications of Trigonometry in real life?

As trigonometry involves the relationship between the sides and angles, finding the height, and calculating the distance, it has a wide range of applications in marine biology, navigation, aviation department, etc. It is also used in solving major mathematical calculations such as Calculus and Algebra. Vedantu provides a solution guide suitable for all the students with solutions that are verified by experienced experts. The solutions are available on both Vedantu’s website and its app at free of cost.

NCERT Solutions for Class 11 Maths

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What is a Trigonometry Table? and How to Create it?

trigonometry questions ncert

  • Updated on  
  • May 20, 2024

Trigonometry Table

A trigonometry table is a reference chart for trigonometry , a branch of math studying angles and triangles. It lists the values of sine (sin), cosine (cos), tangent (tan), and other trigonometric functions for different angles, from 0° to 360°.

Before calculators, these tables were important for solving problems in navigation, science, and engineering. They were so important that they even led to the creation of the first mechanical calculators!

Trigonometry tables are especially useful for right triangles (triangles with a 90° angle). By looking up the angle and the relevant function (sin, cos, etc.), you can find the ratio of the triangle’s sides. This makes solving many geometrical problems much easier.

While calculators have made them less common, trigonometry tables are still a valuable tool for understanding these functions and their applications.Let us read this article to clear your doubts regarding trigonometry table.

Table of Contents

  • 1 What is a Trigonometry Table?
  • 2 How to Remember a Trigonometry Table?
  • 3 How to Create a Trigonometry Table?

What is a Trigonometry Table?

Tables of trigonometric functions are helpful in many different areas of mathematics. Prior to the invention of pocket calculators, trigonometric tables were important for scientific research, engineering, and navigation.Let’s have a look at the specific angle value-based trigonometry formulas:

How to Remember a Trigonometry Table?

Don’t worry about memorizing the trigonometry table! With a few important formulas, you can learn the table in no time.

These formulas relate the trigonometric values (sine, cosine, tangent) of one angle to another. The magic number? 90 degrees!

Here’s the trick to solve such questions based on trigonometry table :

  • To find the sine, cosine, or tangent of any angle, subtract it from 90 degrees and look up that value in your table. The corresponding sine becomes your cosine, cosine becomes sine, and tangent becomes cotangent (and vice versa).
  • By memorising these relationships and just a few table values, you can derive the rest! This technique is much easier than rote memorization and opens the door to solving many trigonometry problems. 

Must Read: 21 Trignometry Formulas for Competitive Exams

How to Create a Trigonometry Table?

These steps provide a systematic approach to filling out a trigonometry table with values for various angles and trigonometric functions.

Step 1: Setting up the Table

  • Create a table with angles listed in the top row, such as 0°, 30°, 45°, 60°, 90°, etc.
  • In the first column, list all trigonometric functions: sin, cos, tan, cosec, sec, cot.

Step 2: Determining sin Values

  • For each angle, divide 0, 1, 2, 3, 4 by 4 under the root, respectively.
  • For example, sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1.

Step 3: Determining cos Values

  • Cos-values are opposite to sin-values. Divide by 4 in the opposite sequence of sin.
  • For example, cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0.

Step 4: Determining tan Values

  • Tan is sin divided by cos (tan = sin/cos).
  • For example, tan 0° = 0/1 = 0, tan 30° = (1/2) / (√3/2) = 1/√3, tan 45° = 1/(1/√2) = √2, and so on.

Step 5: Determining cot Values

  • Cot is the reciprocal of tan.
  • For example, cot 0° = 1/0 = ∞, cot 30° = 1/(1/√3) = √3, cot 45° = 1/1 = 1, and so on.

Step 6: Determining cosec Values

  • Cosec is the reciprocal of sin.
  • For example, cosec 0° = 1/0 = ∞, cosec 30° = 1/(1/2) = 2, cosec 45° = 1/(1/√2) = √2, and so on.

Step 7: Determining sec Values

  • Sec is the reciprocal of cos.
  • For example, sec 0° = 1/1 = 1, sec 30° = 1/(√3/2) = 2/√3, sec 45° = 1/(1/√2) = √2, and so on.

Related Posts

Seven basic trigonometric formulas: Sine, Cosine, Tangent, Cosecant, Secant, Cotangent, and their reciprocal relationships.

Trigonometry has numerous formulas, including trigonometric identities, angle addition, double angle, half angle, and Pythagorean identities.

Six fundamental trigonometric formulas: Sine, Cosine, Tangent, Cosecant, Secant, Cotangent, representing ratios of sides in right triangles.

A trigonometry value table lists angles alongside their corresponding trigonometric function values: sine, cosine, tangent, cosecant, secant, and cotangent.

This was all about the “ Trigonometry Table ”. For more such informative blogs, check out our Study Material Section , or you can learn more about us by visiting our Indian exams page.

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  • School Guide
  • Class 11 Syllabus
  • Class 11 Revision Notes
  • Maths Notes Class 11
  • Physics Notes Class 11
  • Chemistry Notes Class 11
  • Biology Notes Class 11
  • NCERT Solutions Class 11 Maths
  • RD Sharma Solutions Class 11
  • Math Formulas Class 11
  • NCERT Solutions For Class 8 Political Science Chapter-1 The Constitution
  • Chapter 2: Secularism - NCERT Solutions For Class 8 Political Science
  • NCERT Solutions for Class 9 Political Science (Civics)
  • NCERT Solutions Class 9 Civics Social Science Chapter 3: Electoral Politics
  • NCERT Solutions for Class 10 Political Science Chapter 8 : Challenges to Democracy
  • Chapter 8: Law and Social Justice - NCERT Solutions For Class 8 Political Science
  • NCERT Solutions for Class 10 Science Chapter-13 Our Environment
  • NCERT Solutions Class 9 Geography Social Science Chapter 6: Population
  • Legislature| Chapter 5 Class 11 Polity Notes
  • NCERT Solutions Class 9 History Chapter 1 French Revolution
  • NCERT Solutions Class 8 Geography Social Science Chapter 4 Industries
  • NCERT Solutions For Class 10 Political Science (Civics) Chapter 4: Gender, Religion and Caste
  • Chapter 7: Public Facilities NCERT Solutions For Class 8
  • NCERT Solutions For Class 9 Geography Social Science Chapter 4: Climate
  • NCERT Solutions Class 9 Economics Social Science Chapter 2: People as Resource
  • NCERT Solutions Class 8 Geography Chapter 1 Resources
  • NCERT Solution for Class 8 Science Chapter 6 Reproduction in Animals
  • NCERT Solutions For Class 10 Political Science (Civics) Social Science Chapter 1 : Power-sharing
  • NCERT Solutions for Class 10 Science Chapter 8 Heredity

NCERT Solutions Class-11 Political Science Chapter-5: Legislature

NCERT Solutions Class 11 Political Science Chapter 5 Legislature  – This article includes the free NCERT Solutions for Class 11 Polity Chapter 5 Legislature. It will help the students of Class 11 to learn the solutions and ace their exams. It has been developed by the subject matter experts at GFG, according to the  latest CBSE Syllabus 2023-24 , and guidelines. It helps the students of Class 11 create a solid conceptual base for Polity Chapter 5 Legislature.

The solutions to all the exercises in Class 11 Polity Chapter 5 Legislature of your NCERT textbook have been collectively covered in  NCERT Solution for Class 11 (2024-2025) .

NCERT Solutions for Class 11 Polity Chapter 5: Legislature

Page – 107, q. do you think that the composition of the rajya sabha has protected the position of the states of india.

The composition of the Rajya Sabha represents India from different states. It ensures the balance between the interests of the states and central government as well. The Lok Sabha members are elected by the people and Rajya Sabha members are elected by the State Legislative Assemblies.

Q. Should indirect election of Rajya Sabha be replaced by direct elections? What would be its advantages and disadvantages?

The direct elections of the Rajya Sabha are an ongoing debate in India. It enhances accountability to the electorate which leads to increased responsiveness. It also makes democracy legitimate and reduces the political influence in the democracy.

Q. Since 1971 census the number of seats in the Lok Sabha has not increased. Do you think that it should be increased? What should be the basis for this?

It is a complex one because it creates a problem with the representation. India’s population has grown significantly but the Lok Sabha seats have not increased at all. It is based on a careful consideration process as it relates to federalism as well.

Page – 114

Q. from the discussion of the law making process, do you think that parliament can devote enough time for thorough discussion of the bills if not, then what remedies would you suggest to overcome this difficulty.

In the law-making process, the Parliament devotes enough time in the process of discussing bills as it is a matter of concern. There are so many processes and remedies that can be conducted by the comprehensive discussion process to improve the process of equality and effectiveness as well.

Exercise Questions

Q1. alok thinks that a country needs an efficient government that looks after the welfare of the people. so, if we simply elected our prime minister and ministers and left to them the task of government, we will not need a legislature. do you agree give reasons for your answer..

No, I do not agree with Alok as the leaving whole task of the government with a set of ministers will hamper the process and order of democracy. The decisions made by the cabinet would not be accountable as ministers are not bound to explain their decisions to the legislature.

Q2. A class was debating the merits of a bicameral system. The following points were made during the discussion. Read the arguments and say if you agree or disagree with each of them, giving reasons.

(a) neha said that bicameral legislature does not serve any purpose., (b) shama argued that experts should be nominated in the second chamber., (c) tridib said that if a country is not a federation, then there is no need to have a second chamber..

(a) I disagree with Neha’s view because bicameral legislature serves many purposes. It helps in representation to all the sections of the society and a bicameral legislature. (b) I agree with Shama that the experts should be nominated in the second chamber because the first chamber is the House of Representatives where members are elected by the people. c) I disagree with this because even in a unitary state a second chamber is required.

Q3. Why can the Lok Sabha control the executive more effectively than the Rajya Sabha can?

In a Parliamentary government, the executive is responsible to the legislature. As per the constitution, the council of Ministers makes the Lok Sabha more responsible.

Q4. Rather than effective control of the executive, the Lok Sabha is a platform for the expression of popular sentiments and people’s expectations. Do you agree? Give reasons.

Yes, I completely agree with the above statement as control of the Executive is much more effective than others. The representative of the people expresses the difficulties of the people and draws the attention of the government to those problems.

Q5. Arif wanted to know that if ministers propose most of the important bills and if the majority party often gets the government bills passed, what is the role of the Parliament in the law making process? What answer would you give him?

Yes, Ministers proposes the so many important bills in the presence of the majority party. The bill undergoes so many stages where it gets approval and rejections afterwards. At the first stage the general discussion takes place when the bil is done, The committe stage the bill and discuss it with its clauses. At the report stage, the members of the parliament only discuss the bill where they check the need of ammendment of the bill.

Q6. Which of the following statements you agree with the most? Give your reasons.

(a) legislators must be free to join any party they want., (b) anti-defection law has contributed to the domination of the party leaders over the legislators., (c) defection is always for selfish purposes and therefore, a legislator who wants to join another party must be disqualified from being a minister for the next two years..

(a) I do not agree with the above statement. I believe that the legislators should not be free to join any party because this could lead to corruption. (b) It is incorrect as a statement that the anti-defection law always led to the domination of party leaders over legislatures. (c) I do not agree with the above situation because defection is always not for selfish purpose as a leader may use defection to show their disapproval.

Q7. Arrange the different stages of passing of a bill into a law in their correct sequence:

(a) a resolution is passed to admit the bill for discussion., (b) the bill is referred to the president of india – write what happens next if s/he does not sign it., (c) the bill is referred to other house and is passed., (d) the bill is passed in the house in which it was proposed., (e) the bill is read clause by clause and each is voted upon., (f) the bill is referred to the subcommittee – the committee makes some changes and sends it back to the house for discussion., (g) the concerned minister proposes the need for a bill., (h) legislative department in ministry of law, drafts a bill..

(a) A resolution is passed to admit the bill for discussion. (b) The bill is referred to the President of India – write what happens next if s/he does not sign it. (c) The bill is referred to the other House and is passed. (d) The bill is passed in the house in which it was proposed. (e) The bill is read clause by clause and each is voted upon. (f) The bill is referred to the subcommittee – the committee makes some changes and sends it back to the house for discussion. (g) The concerned minister proposes the need for a bill. (h) Legislative department in ministry of law, drafts a bill.

Q8. How has the system of parliamentary committee affected the overseeing and appraisal of legislation by the Parliament?

Parliamentary committee has played a very crucial role in law making since 1983. India’s parliamentary standing committee has reduced the burden of the Parliament. It examines the bill thoroughly and in detail clause by clause and the committee is to submit its report to the house within the prescribe period.

People Also Read:

  • State Legislature
  • Separation of Powers in India
  • Other Forms of Government: Forms and Composition

FAQs on NCERT Solutions Class 11 Polity Chapter 5: Legislature

Who are called legislative.

Legislature of the Union, which is called Parliament, consists of the President and two Houses, known as Council of States (Rajya Sabha) and House of the People (Lok Sabha). Each House has to meet within six months of its previous sitting.

What do you mean by legislation?

legislation, the preparing and enacting of laws by local, state, or national legislatures. In other contexts it is sometimes used to apply to municipal ordinances and to the rules and regulations of administrative agencies passed in the exercise of delegated legislative functions.

What is the work of the legislature?

The Legislature would make laws. The function of the legislative is to form policies and laws that will govern the nation. The Executive implements those laws. The Judiciary checks on the entire process.

Who is the leader of legislature?

The President of India, in their role as head of the legislature, has full powers to summon and prorogue either house of Parliament or to dissolve the Lok Sabha, but they can exercise these powers only upon the advice of the Prime Minister and their Union Council of Ministers.

What are the features of the legislature?

Their powers may include passing laws, establishing the government’s budget, confirming executive appointments, ratifying treaties, investigating the executive branch, impeaching and removing from office members of the executive and judiciary, and redressing constituents’ grievances.

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