Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 627: 72


$\dfrac{3}{\sqrt {x^2-9}}$

Work Step by Step

Suppose $\theta =\sin^{-1} (\dfrac{\sqrt {x^2-9}}{x})$ This gives: $\sin \theta=\dfrac{\sqrt {x^2-9}}{x}$ Since, $ r=\sqrt {x^2+y^2}$ We know $ a=\sqrt {x^2-(x^2-9)}=3$ Therefore, we have $\cot [\sin^{-1} (\dfrac{\sqrt {x^2-9}}{x})]=\dfrac{3}{\sqrt {x^2-9}}$
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