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: 86

Answer

$\tan 2\theta=\frac{-2\tan\theta}{\sec^2\theta-2}$

Work Step by Step

Start with the right side: $\frac{-2\tan\theta}{\sec^2\theta-2}$ Rewrite $\sec^2\theta$ as $\tan^2\theta+1$: $=\frac{-2\tan\theta}{\tan^2\theta+1-2}$ $=\frac{-2\tan\theta}{\tan^2\theta-1}$ Multiply top and bottom by $-1$: $=\frac{-2\tan\theta}{\tan^2\theta-1}*\frac{-1}{-1}$ $=\frac{2\tan\theta}{1-\tan^2\theta}$ Use the identity $\frac{2\tan\theta}{1-\tan^2\theta}=\tan 2\theta$: $=\tan 2\theta$ Since this equals the left side, the identity has been proven.
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