# 2 tan 30°1−tan2 30° is equal to

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 $(c)$