Choose the correct option and justify your choice:
(i) A
(ii) D
(iii) A
(iv) C
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 \(\tan 60^\circ\).