Answer
$-\sqrt {15}$
Work Step by Step
1. Let $g^{-1}(-\frac{1}{4})=t$, we have $g(t)=cos(t)=-\frac{1}{4}$.
2. Let $r=4,x=-1$, we have $y=\sqrt {4^2-1}=\sqrt {15}$
3. Thus $h(g^{-1}(-\frac{1}{4}))=tan(t)=\frac{y}{x}=-\sqrt {15}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 481: 77
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