Answer
$\frac{3}{4}$
Work Step by Step
Within the restricted interval ($0\lt x\lt1$),
$\tan(\sin^{-1}x)=\frac{x}{\sqrt {1-x^{2}}}$
$\implies \tan(\sin^{-1}\frac{3}{5})=\frac{\frac{3}{5}}{\sqrt {1-(\frac{3}{5})^{2}}}=\frac{3}{4}$
Trigonometry 7th Edition
by McKeague, Charles P.; Turner, Mark D.
Published by
Cengage Learning
ISBN 10:
1111826854
ISBN 13:
978-1-11182-685-7
Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - 4.7 Problem Set - Page 262: 73
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