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 693: 28


$=\dfrac {1}{4}\tan 68$

Work Step by Step

$\dfrac {2\tan \alpha }{1-\tan ^{2}\alpha }=\tan 2\alpha \Rightarrow \dfrac {\tan 34}{2\left( 1-\tan ^{2}34\right) }=\dfrac {1}{4}\dfrac {2\tan 34}{1-\tan ^{2}34}=\dfrac {1}{4}\tan 68$
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