Trigonometry (11th Edition) Clone

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

Chapter 6 - Review Exercises - Page 291: 35


$tan~(arcsec\frac{\sqrt{u^2+1}}{u}) = \frac{1}{u}$

Work Step by Step

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