Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Summary Exercises Verifying Trigonometric Identities - Exercises - Page 696: 12

Answer

$1-\tan ^{2}\dfrac {\theta }{2}=\dfrac {2\cos \theta }{1+\cos \theta }$

Work Step by Step

$1-\tan ^{2}\dfrac {\theta }{2}=1-\dfrac {\sin ^{2}\dfrac {\theta }{2}}{\cos ^{2}\dfrac {\theta }{2}}=\dfrac {\cos ^{2}\dfrac {\theta }{2}-\sin ^{2}\dfrac {\theta }{2}}{\cos ^{2}\dfrac {\theta }{2}}=\dfrac {\cos \left( 2\times \dfrac {\theta }{2}\right) }{\dfrac {1+\cos \theta }{2}}=\dfrac {2\cos \theta }{1+\cos \theta }$
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