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

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\)
(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
(v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)
Solution :
(i) sin 60° cos 30° + sin 30° cos 60°
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
= \(\frac{3}{4}+\frac{1}{4}\) = 1
Hence, sin 60° cos 30° + sin 30° cos 60° = 1.

Haryana Board Solutions for 10th Class Maths Chapter 8 Introduction to Trigonometry Ex 8.2

(ii)2 tan2 45° + cos2 30° – sin2 60°
= 2 × (1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)
= 2 × 1 + \(\frac{3}{4}-\frac{3}{4}\) = 2
Hence, 2 tan2 45° + cos2 30° – sin2 60° = 2.

(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\)

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 1

Hence, \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\) = \(\frac{1}{8}\) (3√2 – √6).

(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 2

Hence, \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\) = \(\frac{43-24 \sqrt{3}}{11}\)

(v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 3

Haryana Board Solutions for 10th Class Maths Chapter 8 Introduction to Trigonometry Ex 8.2

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°

Haryana Board Solutions for 10th Class Maths Chapter 8 Introduction to Trigonometry Ex 8.2
Solution:
(i) (a)

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 4

(ii) (d)
∵ \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}\)
= \(\frac{1-1}{1+1}=\frac{0}{2}\) = 0

(iii) (a) When A = 0°
L.H.S. = sin 2A = sin 2 × 0°
= sin 0° = 0
R.H.S. = 2 sin A = 2 × sin 0°
= 2 × 0 = 0
In other option we shall find that L.H.S. ≠ R.H.S.

(iv) (c)
∵ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\)
= \(\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}\)
= √3 = tan 60°.

Haryana Board Solutions for 10th Class Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Question 3.
If tan(A + B) = √3 and tan(A – B) = \(\frac{1}{\sqrt{3}}\); 0° < (A + B) ≤ 90°, A > B, find A and B.
Solution :
We have
tan(A + B) = √3
tan(A + B) = tan 60°
A + B = 60° ……………..(1)
and tan(A – B) = \(\frac{1}{\sqrt{3}}\)
tan(A – B) = tan 30°
A – B = 30° ……………….(2)
Adding equation (1) and (2), we get
A + B = 60°
A – B = 30°
2A = 90°
A = \(\frac{90^{\circ}}{2}\)
A = 45°
Substituting the value of A in equation (1), we get
45° + B = 60°
B = 60° – 45° = 15°
Hence, A = 45° and B = 15°.

Haryana Board Solutions for 10th Class Maths Chapter 8 Introduction to Trigonometry Ex 8.2

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 a 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. Taking A= 30°, B = 60°
LHS = sin (A + B)
= sin (30° + 60°)
= sin 90° = 1
RHS = sin A + sin B
= sin 30° + sin 60°
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\)
∴ LHS ≠ RHS
(ii) True.
(iii) False. Because value of cos θ decreases as θ increases.
(iv) False. Because sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
i.e., sin 30° ≠ cos 30°
(v) True. Because cot 0° = ∞ (not defined)

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