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: 22


See the explanation below.

Work Step by Step

$\frac{{{\cot }^{2}}t}{\csc t}=\csc t-\sin t$ Recall Trigonometric Identities, $\begin{align} & 1+{{\cot }^{2}}t={{\csc }^{2}}t \\ & \csc t=\frac{1}{\sin t} \\ \end{align}$ Consider the left side of the given expression, $\begin{align} & \frac{{{\cot }^{2}}t}{\csc t}=\frac{{{\csc }^{2}}t-1}{\csc t} \\ & =\csc t-\frac{1}{\csc t} \\ & =\csc t-\sin t \end{align}$ Therefore, $\frac{{{\cot }^{2}}t}{\csc t}=\csc t-\sin t$
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