Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises: 8

Answer

$-\cos^{2}\theta\cdot\csc\theta$

Work Step by Step

$\cot(-\theta)=-\cot\theta \quad \sec(-\theta)=\sec\theta .$ $\displaystyle \cot\theta=\frac{\cos\theta}{\sin\theta}, \displaystyle \quad \sec\theta=\frac{1}{\cos\theta}\quad \csc\theta=\frac{1}{\sin\theta}$ -------------- $\displaystyle \frac{\cot(-\theta)}{\sec(-\theta)}=\frac{-\cot\theta}{\sec\theta}=-\cot\theta\times\frac{1}{\sec\theta}$ $=-\displaystyle \frac{\cos\theta}{\sin\theta}\times\frac{1}{\sec\theta}$ $=-\displaystyle \frac{\cos\theta}{\sin\theta}\times\cos\theta$ $=-\displaystyle \cos^{2}\theta\times\frac{1}{\sin\theta}$ $=-\cos^{2}\theta\cdot\csc\theta$
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