Answer
$\dfrac{\pi}{6}$
Work Step by Step
$\cot^{-1}{(\sqrt3)}=\dfrac{\pi}{6}$, because $\cot{\left(\dfrac{\pi}{6}\right)}=\sqrt3$ and $\dfrac{\pi}{6}$ is in the range of $\cot^{-1}{x}$, which is $(0,\pi).$
Precalculus (10th Edition)
by Sullivan, Michael
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: 37
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