Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises - Page 659: 65

Answer

$\sin ^{2}\theta \cos ^{2}\theta $

Work Step by Step

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