Answer
$$\frac{2\tan\frac{\pi}{3}}{1-\tan^2\frac{\pi}{3}}=-\sqrt3$$
6 would be matched with D.
Work Step by Step
$$X=\frac{2\tan\frac{\pi}{3}}{1-\tan^2\frac{\pi}{3}}$$
From the double-angle identities:
$$\frac{2\tan A}{1-\tan^2 A}=\tan 2A$$
Thus $\frac{2\tan\frac{\pi}{3}}{1-\tan^2\frac{\pi}{3}}$ can be seen here as $\frac{2\tan A}{1-\tan^2 A}$ with $A=\frac{\pi}{3}$.
Therefore,
$$X=\tan\Big(2\times\frac{\pi}{3}\Big)$$
$$X=\tan\frac{2\pi}{3}$$
$|\tan\frac{2\pi}{3}|=\tan\frac{\pi}{3}$. As $\frac{2\pi}{3}$ is in quadrant II, in which $\sin\theta\gt0$ but $\cos\theta\lt0$, so $\tan\theta=\frac{\sin\theta}{\cos\theta}\lt0$. Therefore, $\tan\frac{2\pi}{3}=-\tan\frac{\pi}{3}=-\sqrt3$.
$$X=-\sqrt3$$
So, $$\frac{2\tan\frac{\pi}{3}}{1-\tan^2\frac{\pi}{3}}=-\sqrt3$$
6 would be matched with D.
