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: 39

Answer

$-\dfrac{\pi}{2}$

Work Step by Step

$\csc^{-1}{(-1)}=-\dfrac{\pi}{2}$, because $\csc{\left(-\dfrac{\pi}{2}\right)}=-1$ and $-\dfrac{\pi}{2}$ is in the range of $\csc^{-1}{x}$, which is $\left[-\dfrac{\pi}{2}, 0\right) \cup \left(0, \dfrac{\pi}{2}\right].$
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