Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Mid-Chapter Check Point - Page 684: 14


Please see below.

Work Step by Step

We verify as follows: $$\frac{\sec ^2x}{2- \sec ^2 x}=\frac{1}{\frac{2}{\sec ^2 x}-1}=\frac{1}{2\cos ^2 x-1}=\frac{1}{\cos 2x}=\sec 2x$$
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