## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 18

#### Answer

$\displaystyle \frac{\pi}{2}$

#### Work Step by Step

$y=\sec^{-1}x$ is the number in $[0, \pi/2) \cup(\pi/2, \pi$] for which $\sec y=x.$ (In terms of cosine, $\displaystyle \qquad \cos y=\frac{1}{x}$) When cosine approaches zero from the left, the secant function falls toward $-\infty.$ This angle (in radians) is $\displaystyle \frac{\pi}{2}$ (we observe $[0, \pi/2) \cup(\pi/2, \pi]$). Alternatively (if in doubt), we can reach the same conclusion by observing the graph of $y=\sec x$ (also written as $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{e}\mathrm{c} x$) when $x\rightarrow-\infty$. See below.

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.