## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Mid-Chapter Check Point - Page 577: 11

#### Answer

$$\cos \theta = -\frac{4}{5} \\ \sec \theta = -\frac{5}{4} \\ \cot \theta =-\frac{4}{3} \\ \sin \theta = \frac{3}{5} \\ \csc \theta = \frac{5}{3}$$

#### Work Step by Step

By using the assumption that $\tan \theta =-\frac{3}{4}, \quad \cos \theta \lt 0$ and applying appropriate trigonometric identities, we can find the value of the remaining trigonometric functions of $\theta$ as follows.$$\cos \theta =\frac{1}{ \sqrt{1+\tan^2 \theta }}= \frac{1}{\sqrt{1+(-\frac{3}{4})^2}}=-\frac{4}{5} \\ \sec \theta = \frac{1}{\cos \theta }=\frac{1}{-\frac{4}{5}}=-\frac{5}{4} \\ \cot \theta =\frac{1}{\tan \theta }= \frac{1}{-\frac{3}{4}}=-\frac{4}{3} \\ \sin \theta = (\tan \theta )(\cos \theta )=(-\frac{3}{4})(-\frac{4}{5})=\frac{3}{5} \\ \csc \theta = \frac{1}{\sin \theta }= \frac{1}{\frac{3}{5}}=\frac{5}{3}$$

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.