Precalculus (6th Edition)

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


$ 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$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.