# 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) $\sin 30^{\circ}$.

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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,

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

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

$=0$

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

LHS $=\sin 2 A$

$=\sin 2 \times 0$

$=\sin 0^{\circ}$

$=0$

RHS $=2 \sin A$

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

$=2 \times 0$

$=0$

(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.

Updated on 10-Oct-2022 13:22:12