Trigonometry (11th Edition) Clone

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

Chapter 6 - Test - Page 293: 8


$tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$

Work Step by Step

$\theta = arcsin(u)$ $sin~\theta = u = \frac{u}{1} = \frac{opposite}{hypotenuse}$ Note that $\theta$ is in quadrant I since $u \gt 0$. We can find the value of the adjacent side: $adjacent = \sqrt{1^2-u^2}$ We can find the value of $tan~\theta$: $tan~\theta = \frac{opposite}{adjacent}$ $tan~\theta = \frac{u}{\sqrt{1^2-u^2}}$ $tan~\theta = \frac{u}{\sqrt{1^2-u^2}}~\frac{\sqrt{1^2-u^2}}{\sqrt{1^2-u^2}}$ $tan~\theta = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$ Therefore, $tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$
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