Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.4 Trigonometric Identities - 7.4 Assess Your Understanding - Page 476: 60

Answer

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Work Step by Step

$LHS=\frac{sin\theta cos\theta}{cos^2\theta-sin^2\theta}\times\frac{1/cos^2\theta}{1/cos^2\theta}=\frac{sin\theta/cos\theta}{1-sin^2\theta/cos^2\theta}=\frac{tan\theta}{1-tan^2\theta}=RHS$
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