NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY - Exercise 8.2

Question 2

We solve each part using standard trigonometric identities and exact values of special angles.

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

Identity used: \(\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta}\).

Here \(\theta = 30^\circ\). So,

\(\dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \sin(2\times 30^\circ) = \sin 60^\circ\).

Correct option: (A) \(\sin 60^\circ\).

Student Note: Whenever you see \(\dfrac{2\tan\theta}{1+\tan^2\theta}\), it directly matches \(\sin 2\theta\).

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

We know \(\tan 45^\circ = 1\), so \(\tan^2 45^\circ = 1\).

Substitute:

Numerator: \(1 - 1 = 0\)

Denominator: \(1 + 1 = 2\)

Therefore, \(\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \dfrac{0}{2} = 0\).

Correct option: (D) 0.

(iii) “\(\sin 2A = 2\sin A\)” is true when \(A = ?\)

Identity used: \(\sin 2A = 2\sin A\cos A\).

Given equation:

\(2\sin A\cos A = 2\sin A\)

Bring to one side:

\(2\sin A(\cos A - 1) = 0\)

So, either:

(1) \(\sin A = 0\) or (2) \(\cos A - 1 = 0\Rightarrow \cos A = 1\).

For the given options, \(A = 0^\circ\) makes \(\sin A = 0\) and \(\cos A = 1\), so the equation is true.

Correct option: (A) \(0^\circ\).

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

Identity used: \(\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}\).

Here \(\theta = 30^\circ\). So,

\(\dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \tan(2\times 30^\circ) = \tan 60^\circ\).

Correct option: (C) \(\tan 60^\circ\).

Quick Check: \(\tan 60^\circ = \sqrt{3}\), so the expression should come out greater than 1, which matches \(\tan 60^\circ\).