Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 38

Answer

$60^o$

Work Step by Step

The value of $\theta$ must be in the interval $(-90^o, 90^o)$. Note that $\tan{60^o} = \sqrt{3}$. Thus, $\tan^{-1}{\sqrt3}=60^o$
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