Answer
$θ = -\frac{\pi}{6}$
Work Step by Step
$arccot(-\sqrt{3})=arccot(-\frac{\sqrt{3}}{1})$
$=\arctan(-\frac{1}{\sqrt{3}}=\arctan(-\frac{\sqrt{3}}{3})$
$=\tan^{-1}(-\frac{-\sqrt{3}}{3})=θ$
$\tan θ=-\frac{-\sqrt{3}}{3}$
$θ = -\frac{\pi}{6}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 372: 8
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