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: 22

Answer

$-\dfrac{\pi}{6}$

Work Step by Step

Note that $\cot{\left(\dfrac{2\pi}{3}\right)}=-\dfrac{\sqrt3}{3}$. Thus $\tan^{-1}\left(\cot{\left(\dfrac{2\pi}{3}\right)}\right)=\tan^{-1}{\left(-\dfrac{\sqrt3}{3}\right)}=-\dfrac{\pi}{6}$, because $\tan{\left(-\dfrac{\pi}{6}\right)}=-\dfrac{\sqrt3}{3}$ and $-\dfrac{\pi}{6}$ is in the range of $\tan^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$
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