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: 103


$sec~(arccot~\frac{\sqrt{4-u^2}}{u}) = \frac{2~\sqrt{4-u^2}}{4-u^2}$

Work Step by Step

Let $~~\theta = arccot~\frac{\sqrt{4-u^2}}{u}$ Then $~~cot~\theta = \frac{\sqrt{4-u^2}}{u}$ We can find an expression for $sec~\theta$: $sec~\theta = \frac{\sqrt{(\sqrt{4-u^2})^2+u^2}}{\sqrt{4-u^2}}$ $sec~\theta = \frac{\sqrt{4-u^2+u^2}}{\sqrt{4-u^2}}$ $sec~\theta = \frac{\sqrt{4}}{\sqrt{4-u^2}}$ $sec~\theta = \frac{2}{\sqrt{4-u^2}}$ $sec~\theta = (\frac{2}{\sqrt{4-u^2}})~(\frac{\sqrt{4-u^2}}{\sqrt{4-u^2}})$ $sec~\theta = \frac{2~\sqrt{4-u^2}}{4-u^2}$ Therefore, $~~sec~(arccot~\frac{\sqrt{4-u^2}}{u}) = \frac{2~\sqrt{4-u^2}}{4-u^2}$
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