Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 34

Answer

$ \sqrt 5$

Work Step by Step

Step 1. Letting $tan^{-1}(\frac{1}{2})=t$, we have $tan(t)=\frac{1}{2}$ and $0\lt t\lt\frac{\pi}{2}$. Step 2. Let $x=2, y=1$, and $r=\sqrt {x^2+y^2}=\sqrt {5}$. Step 3. Sides $x,y,r$ form a right triangle with angle $|t|$ facing $y$. Step 4. We know $csc(t)$ is positive and we have $csc(t)=\frac{r}{y}=\sqrt 5$ or $csc(tan^{-1}(\frac{1}{2}))= \sqrt 5$
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