# Choose the correct option and justify your choice:

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Question.

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) $\sin 2 \mathrm{~A}=2 \sin \mathrm{A}$ is true when $\mathrm{A}=$

(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:

(i) Option (A) is correct

$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$

$=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$

(ii) Option (D) is correct.

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-1}{1+1}=0$

(iii) Option (A) is correct.

sin 2A = 2sinA

2sinA . cosA = 2sinA

cosA = 1

A = 0°

(iv) Option (C) is correct

$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$

$=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$

$=\tan 60^{\circ}$