Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 658: 21


See the explanation below.

Work Step by Step

$\frac{{{\tan }^{2}}t}{\sec t}=\sec t-\cos t$ Recall Trigonometric Identities, $\begin{align} & 1+{{\tan }^{2}}t={{\sec }^{2}}t \\ & \sec t=\frac{1}{\cos t} \\ \end{align}$ Use the above identities and solve the left side of the given expression, $\begin{align} & \frac{{{\tan }^{2}}t}{\sec t}=\frac{{{\sec }^{2}}t-1}{\sec t} \\ & =\sec t-\frac{1}{\sec t} \\ & =\sec t-\cos t \end{align}$ Therefore, $\frac{{{\tan }^{2}}t}{\sec t}=\sec t-\cos t$
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