Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.7 Derivatives And Integrals Involving Inverse Trigonometric Functions - Exercises Set 6.7 - Page 472: 85

Answer

(a)$sin^{-1}(-x)=-sin^{-1}x$ (b) $tan^{-1}(-x)=-tan^{-1}x$

Work Step by Step

(a)Let $\sin^{-1}(-x)=y$ ................. eq (1) $\Longrightarrow$ $-x=siny$ Or $x=-siny=sin(-y)$ Now $x=sin(-y)$ $ \Longrightarrow$ $sin^{-1}x=-y$ Or $-sin^{-1}x=y$ .............. eq (2) From equation (1) and equation (2) $sin^{-1}(-x)=-sin^{-1}x$ (b)Let $tan^{-1}(-x)=y$ ...................... eq (3) $\Longrightarrow$ $tany=-x$ Or $-tany=x$ Or $tan(-y)=x$ $tan^{-1}x=-y$ Or $-tan^{-1}x=y$ .................... eq (4) From equation (3) and equation (4) $tan^{-1}(-x)=-tan^{-1}x$
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