# NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry Exercise 8.2

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### Access other exercise solutions of Class 10 Maths Chapter 8- Introduction to Trigonometry

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

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)

### Access Answers of Maths NCERT Class 10 Chapter 8 – Introduction to Trigonometry Exercise 8.2

1. Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan2 45° + cos2 30° – sin2 60

Solution:

(i) sin 60° cos 30° + sin 30° cos 60°

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

Now, substitute the values in the given problem

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

(ii) 2 tan2 45° + cos2 30° – sin2 60

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

sin 60° = √3/2

cos 30° = √3/2

tan 45° = 1

Substitute the values in the given problem

2 tan2 45° + cos2 30° – sin2 60 = 2(1)2 + (√3/2)2-(√3/2)2

2 tan2 45° + cos2 30° – sin2 60 = 2 + 0

2 tan2 45° + cos2 30° – sin2 60 = 2

(iii) cos 45°/(sec 30°+cosec 30°)

We know that,

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

Substitute the values, we get

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

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

We know that,

sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

Substitute the values in the given problem, we get

We know that,

cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

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

(5cos260° + 4sec230° – tan245°)/(sin2 30° + cos2 30°)

= 5(1/2)2+4(2/√3)2-12/(1/2)2+(√3/2)2

= (5/4+16/3-1)/(1/4+3/4)

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

= 67/12

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(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-tan230° =
(A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

Solution:

(i) (A) is correct.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan 30°/1+tan230° = 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

tan 45° = 1

1-tan245°/1+tan245° = (1-12)/(1+12)

= 0/2 = 0

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

or,

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.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan30°/1-tan230° =  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.

Solution:

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°

Now substitute the degree value

⇒ 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

2A = 90°

A= 45°

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

45° + B = 60°

B = 60° – 45°

B = 15°

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°.

Solution:

(i) False.

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.

(ii) True.

Justification:

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

sin 0° = 0

sin 30° = 1/2

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 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

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

(iv) False

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

(v) True.

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 8.2 of Class 10 Maths covers concepts like trigonometric ratios of some specific angles of a right angle triangle like trigonometric ratios of 0°,45° and 90° with the help of right angle triangle and the trigonometric ratios of 30° and 60° with the help of equilateral triangle are also explained.

A trigonometric table is provided with all the angle measures from 0° to 90°, which shows that the increase from 0° to 90°, the value of sine increases from 0 to 1 and the value of cosine decreases from 1 to 0. From these two trigonometric ratio values, the value of other functions can be determined. Since the tangent function is the ratio of sine by cos function and the inverse functions of sin, cos and tan functions are illustrated with suitable examples.

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