Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 62


$\sec^2\alpha-1=\frac{\sec 2\alpha-1}{\sec 2\alpha+1}$

Work Step by Step

Simplify the left side: $\sec^2\alpha-1$ $=\tan^2\alpha$ Simplify the right side: $\frac{\sec 2\alpha-1}{\sec 2\alpha+1}$ Rewrite it in terms of cosine: $=\frac{\frac{1}{\cos 2\alpha}-1}{\frac{1}{\cos 2\alpha}+1}$ Multiply top and bottom by $\cos 2\alpha$: $=\frac{\frac{1}{\cos 2\alpha}-1}{\frac{1}{\cos 2\alpha}+1}*\frac{\cos 2\alpha}{\cos 2\alpha}$ $=\frac{1-\cos 2\alpha}{1+\cos 2\alpha}$ Use the identities $\cos 2\alpha=1-2\sin^2\alpha$ and $\cos 2\alpha=2\cos^2\alpha-1$: $=\frac{1-(1-2\sin^2\alpha)}{1+(2\cos^2\alpha-1)}$ Simplify: $=\frac{1-1+2\sin^2\alpha}{1+2\cos^2\alpha-1}$ $=\frac{2\sin^2\alpha}{2\cos^2\alpha}$ $=\tan^2\alpha$ Since the left side and the right side both simplify to $\tan^2\alpha$, they are equal to each other, and the identity has been proven.
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