In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.

Evaluate the following:

(i) sin60° cos30° + sin 30° cos 60°

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

Solution:

(i) We have

= \(\frac{\frac{1}{12}\times67}{\frac{4}{4}}=\frac{67}{12}\)

Question 2.

Choose the correct option and justify your choice:

(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)

(A) sin 60°

(B) cos 60°

(C) tan 60°

(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)

(A) tan 90°

(B) 1

(C) sin 45°

(D) 0

(iii) sin 2A = 2sin A is true when A =

(A) 0°

(B) 30°

(C) 45°

(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)

(A) cos 60°

(B) sin 60°

(C) tan 60°

(D) sin 30°

Solution:

Question 3.

If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0°< A + B ≤ 90°; A > B, find A and B.

Solution:

We have,

tan 60° = \(\sqrt{3}\), tan 30° = \(\frac{1}{\sqrt{3}}\) ……………. (1)

Also, tan(A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) …………… (2)

From (1) and (2), we get

A + B = 60° …………… (3)

and A – B = 30° ………….. (4)

On adding (3) and (4), we get

2A = 90° ⇒ A =45°

On subtracting (4) from (3), we get

2B = 30° ⇒ B = 15°

Question 4.

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

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

(ii) The value of sin6 increases as θ increases.

(iii) The value of cosθ 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:

Let us take A = 30° and B = 60°, then

L.H.S = sin (30° + 60°) = sin 90° = 1

∴ L.H.S. ≠ R.H.S.

(ii) True:

Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.

(iii) False:

Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.

(iv) False:

Let us take θ = 30°

sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)

⇒ sin 30° ≠ cos 30°

(v) True:

We have, cot 0° = not defined

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.

Evaluate the following:

(i) sin60° cos30° + sin 30° cos 60°

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

Solution:

(i) We have

= \(\frac{\frac{1}{12}\times67}{\frac{4}{4}}=\frac{67}{12}\)

Question 2.

Choose the correct option and justify your choice:

(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)

(A) sin 60°

(B) cos 60°

(C) tan 60°

(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)

(A) tan 90°

(B) 1

(C) sin 45°

(D) 0

(iii) sin 2A = 2sin A is true when A =

(A) 0°

(B) 30°

(C) 45°

(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)

(A) cos 60°

(B) sin 60°

(C) tan 60°

(D) sin 30°

Solution:

Question 3.

If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0°< A + B ≤ 90°; A > B, find A and B.

Solution:

We have,

tan 60° = \(\sqrt{3}\), tan 30° = \(\frac{1}{\sqrt{3}}\) ……………. (1)

Also, tan(A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) …………… (2)

From (1) and (2), we get

A + B = 60° …………… (3)

and A – B = 30° ………….. (4)

On adding (3) and (4), we get

2A = 90° ⇒ A =45°

On subtracting (4) from (3), we get

2B = 30° ⇒ B = 15°

Question 4.

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

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

(ii) The value of sin6 increases as θ increases.

(iii) The value of cosθ 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:

Let us take A = 30° and B = 60°, then

L.H.S = sin (30° + 60°) = sin 90° = 1

∴ L.H.S. ≠ R.H.S.

(ii) True:

Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.

(iii) False:

Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.

(iv) False:

Let us take θ = 30°

sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)

⇒ sin 30° ≠ cos 30°

(v) True:

We have, cot 0° = not defined