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: 17


See the explanation below.

Work Step by Step

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