## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 658: 19

#### Answer

See the explanation below.

#### Work Step by Step

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.