Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.2 Verifying Trigonometric Identities - 7.2 Exercises - Page 667: 72

Answer

$\tan^2\alpha \sin ^{2}\alpha =\tan ^{2}\alpha +\cos ^{2}\alpha -1$

Work Step by Step

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