Answer
$\frac{1}{\sqrt {1+u^2}}$
Work Step by Step
1. Let $tan^{-1}u=t$, we have $tan(t)=u$.
2. Let $x=1,y=u$, we have $r=\sqrt {1+u^2}$
3. Thus $cos(tan^{-1}u)=cos(t)=\frac{x}{r}=\frac{1}{\sqrt {1+u^2}}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 481: 57
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
