#### 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$

Published by
Pearson

ISBN 10:
013421742X

ISBN 13:
978-0-13421-742-0

$ 6\rightarrow D$

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