Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 104


$csc~(arctan~\frac{\sqrt{9-u^2}}{u}) = \frac{3~\sqrt{9-u^2}}{9-u^2}$

Work Step by Step

Let $~~\theta = arctan~\frac{\sqrt{9-u^2}}{u}$ Then $~~tan~\theta = \frac{\sqrt{9-u^2}}{u}$ We can find an expression for $csc~\theta$: $csc~\theta = \frac{\sqrt{(\sqrt{9-u^2})^2+u^2}}{\sqrt{9-u^2}}$ $csc~\theta = \frac{\sqrt{9-u^2+u^2}}{\sqrt{9-u^2}}$ $csc~\theta = \frac{\sqrt{9}}{\sqrt{9-u^2}}$ $csc~\theta = \frac{3}{\sqrt{9-u^2}}$ $csc~\theta = (\frac{3}{\sqrt{9-u^2}})~(\frac{\sqrt{9-u^2}}{\sqrt{9-u^2}})$ $csc~\theta = \frac{3~\sqrt{9-u^2}}{9-u^2}$ Therefore, $~~csc~(arctan~\frac{\sqrt{9-u^2}}{u}) = \frac{3~\sqrt{9-u^2}}{9-u^2}$
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