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$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 692: 6
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