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: 27


$=\dfrac {1}{2}\tan 102$

Work Step by Step

$\dfrac {2\tan \alpha }{1-\tan ^{2}\alpha }=\tan 2\alpha \Rightarrow \dfrac {tun51}{1-\tan ^{2}51}=\dfrac {1}{2}\dfrac {2\tan 51}{1-\tan ^{2}51}=\dfrac {1}{2}\tan 102$
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