Question:
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ is equal to
(a) $\cos 60^{\circ}$
(b) $\sin 60^{\circ}$
(c) $\tan 60^{\circ}$
(d) $\sin 30^{\circ}$
Solution:
We are asked to find the value of the following
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$
$=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$
$=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
We know that $\left[\begin{array}{l}\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \tan 60^{\circ}=\sqrt{3}\end{array}\right]$
$=\frac{3}{\sqrt{3}}$
$=\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=\sqrt{3}$
$=\tan 60^{\circ}$
Hence the correct option is $(c)$