Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.2 - Verifying Trigonometric Identities - 7.2 Exercises - Page 521: 57


The identity is verified. $tan(sin^{-1}x)=\frac{x}{\sqrt {1-x^2}}$

Work Step by Step

$y=sin^{-1}x$, domain: $-1\leq x\leq1$, range: $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$: $x=sin~y$ $x^2=sin^2y$ $1-x^2=1-sin^2y$ $1-x^2=cos^2y$ $cos~y=+\sqrt {1-x^2}~~$ (Use only the $+$ because $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$) $tan(sin^{-1}x)=\frac{sin(sin^{-1}x)}{cos(sin^{-1}x)}=\frac{sin~y}{cos~y}=\frac{x}{\sqrt {1-x^2}}$
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