Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.2

Question 1.

Evaluate the following :

(i) sin 60° cos 30° + sin 30° cos 60°

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

Solution:

(i) According to the quesiton

sin 60° cos 30° + sin 30° cos 60°

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

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

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 = 2 sin 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:

(iii) According to the question sin 2A = 2 sin

When A = 0° then

sin 2(0) = 2 sin 0

⇒ sin 0 = 0

⇒ 0 = 0; which is true.

∴ Correct option = (A)

Correct option = (C)

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:

tan (A + B) = \(\sqrt {3}\)

tan (A + B) = tan 60°

or A + B = 60° …. (i)

tan(A – B) = \(\frac{1}{\sqrt{3}}\)

or tan (A – B) = tan 30°

or A – B = 30° …. (ii)

Adding (i) and (ii)

Putting the value in (i)

A = 45°

45° + B = 60°

B = 60° – 45°

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 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) When A = 60°, B = 30°

L.H.S. = sin (A + B)

= sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A + sin B

= sin 60° + sin 30°

= \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) ≠ 1

i.e., L.H.S. ≠ R.H.S.

This is false.

(ii) sin 0° = 0

sin 30° = \(\frac{1}{2}\) = 0.5

sin 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 (approximately)

sin 60°= \(\frac{\sqrt{3}}{2}\) = 0.87 (approximately)

and sin 90° = 1

i.e., when the value of θ increases from 0° to 90° then the value of sin θ also increases. But it is true upto θ = 90° and not ahead.

∴ This is true.

(iii) cos 0° = 1

cos 30° = \(\frac{\sqrt{3}}{2}\) = 0.87 (approximately)

cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 (approximately)

cos 60°= \(\frac{1}{2}\) = 0.5

and cos 90° = 0

When the value of θ increases from 0° to 90° then the value of cos θ decreases.

Hence it is false.

(iv) sin 30° = \(\frac{1}{2}\)

and cos 30° = \(\frac{\sqrt{3}}{2}\)

or sin 30° ≠ cos 30°

θ = 45°

sin 45° = \(\frac{1}{\sqrt{2}}\)

and cos 45° = \(\frac{1}{\sqrt{2}}\)

only at θ = 45° the values are equal.

But for all values of θ

sin θ ≠ cos θ

∴ The given statement is false.

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