Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 56

Answer

$csc[arctan(-\frac{5}{12})]=-\frac{13}{12}$

Work Step by Step

Let $u=arctan(-\frac{5}{12})$. Then: $tan~u=-\frac{5}{12}$ $cot~u=\frac{1}{tan~u}=-\frac{12}{5}$ The range $arctan~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arctan(-\frac{5}{12})\lt0$, then $-\frac{\pi}{2}\lt u\lt0~~$ (Fourth Quadrant) $csc^2u=1+cot^2u$ $csc^2u=1+(-\frac{5}{12})^2$ $csc^2u=1+\frac{25}{144}=\frac{169}{144}$ $csc~u=-\frac{13}{12}~~$ (Fourth Quadrant) $csc[arctan(-\frac{5}{12})]=csc~u=-\frac{13}{12}$
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