Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 26


$-\sqrt 3$

Work Step by Step

Since $\sec^{-1} x$ is an even function, then we have $\sec^{-1}(-2)=\sec^{-1} 2$. Now, assume that $\sec\theta =\frac{2}{1}$, then by solving the triangle, we have: $$\tan \theta=\sqrt 3.$$ Now, since $\theta =\sec^{-1}2 $, we have $$\tan(\sec^{-1}(-2) )=\tan\theta=-\sqrt 3 .$$ We know that tangent is negative because secant is negative in the 2nd and 3rd quadrants. We exclude the 3rd quadrant because it is outside the range of arcsecant. We know that tangent is negative in the 2nd quadrant, thus the answer must be negative.
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