Answer
$-\sqrt 5$.
Work Step by Step
1. Let $tan^{-1}(-2)=t$, we have $tan(t)=-2$ and $t$ in quadrant IV.
2. Let $x=2, y=-1$, we have $r=\sqrt {2^2+(-1)^2}=\sqrt 5$.
3. Thus $csc(tan^{-1}(-2))=csc(t)=\frac{r}{y}=-\sqrt 5$.
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 480: 30
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