Choose the correct option and justify your choice:
(i) \( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}= \)
(A) \( \sin 60^{\circ} \) (B) \( \cos 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \)
(ii) \( \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}= \)
(A) \( \tan 90^{\circ} \) (B) 1 (C) \( \sin 45^{\circ} \) (D) 0
(iii) \( \sin 2 \mathrm{~A}=2 \sin \mathrm{A} \) is true when \( \mathrm{A}= \)
(A) \( 0^{\circ} \) (B) \( 30^{\circ} \) (C) \( 45^{\circ} \) (D) \( 60^{\circ} \)
(iv) \( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}= \)
(A) \( \cos 60^{\circ} \) (B) \( \sin 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \).

To do:

We have to choose the correct option and justify it.

Solution:  

(i) We know that,

$tan 30^{\circ}=\frac{1}{\sqrt3}$

Therefore,

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

$=\frac{\frac{2}{\sqrt{3}}}{\frac{1}{1}+\frac{1}{3}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}$

$=\frac{2}{\sqrt{3}} \times \frac{3}{4}$

$=\frac{3}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{3 \sqrt{3}}{2 \times 3}$

$=\frac{\sqrt{3}}{2}$

$=\sin 60^{\circ}$

The correct option is A.

(ii) We know that,

$tan\ 45^{\circ}=1$

Therefore,

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

$=\frac{1-1}{1+1}$

$=0$

The correct answer is D.

(iii) If $A=0^{\circ}$, then

LHS $=\sin 2 A$

$=\sin 2 \times 0$

$=\sin 0^{\circ}$

$=0$

RHS $=2 \sin A$

$=2 \sin 0^{\circ}$

$=2 \times 0$

$=0$

The correct answer is A. 

(iv) We know that,

$tan 30^{\circ}=\frac{1}{\sqrt3}$

Therefore,

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

$=\frac{\frac{2}{\sqrt{3}}}{\frac{1}{1}-\frac{1}{3}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}$

$=\frac{2}{\sqrt{3}} \times \frac{3}{2}$

$=\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{3 \sqrt{3}}{3}$

$=\sqrt{3}$

$=\tan 60^{\circ}$

The correct option is C.