Answer
$ 6\rightarrow D$
Work Step by Step
$\dfrac {2\tan \alpha }{1-\tan ^{2}\alpha }=\tan _{2}\alpha \Rightarrow \dfrac {2\tan \dfrac {\pi }{3}}{1-\tan ^{2}\dfrac {\pi }{3}}=\tan \dfrac {2\pi }{3}=-\sqrt {3}\Rightarrow$
$ 6\rightarrow D$