RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

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 tan² 45° + cos² 30° – sin² 60°

Solution:
(i) According to the quesiton
sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 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.