Trigonometry (11th Edition) Clone

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

Chapter 5 - Review Exercises - Page 248: 14


$$\cot x=-\frac{4}{5}$$ $$\sec x=-\frac{\sqrt{41}}{4}$$ $$\csc x=\frac{\sqrt{41}}{5}$$

Work Step by Step

$$\tan x=-\frac{5}{4}\hspace{2cm}\frac{\pi}{2}\lt x\lt \pi$$ 1) Determining the signs of $\cot x$, $\sec x$ and $\csc x$ The place of $x$ is where $\frac{\pi}{2}\lt x\lt \pi$, meaning $x$ is in quadrant II, where sines are positive but cosines are negative. As $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$, this means both tangents and cotangents in quadrant II are negative. As $\sec x=\frac{1}{\cos x}$ and $\csc x=\frac{1}{\sin x}$, this means secants are negative, while cosecants are positive in quadrant II. Therefore, $\cot x\lt0$, $\sec x\lt0$ and $\csc x\gt0$. 2) Calculating $\cot x$ - Recall the Reciprocal Identities: $\cot x=\frac{1}{\tan x}$ Therefore, $$\cot x=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}$$ 3) Calculating $\sec x$ and $\csc x$ - Recall the Pythagorean Identities for secant and cosecants: $$\sec^2x=1+\tan^2x$$ $$\csc^2x=1+\cot^2x$$ Therefore, $$\sec^2x=1+\Big(-\frac{5}{4}\Big)^2=1+\frac{25}{16}=\frac{41}{16}$$ $$\sec x=\pm\frac{\sqrt{41}}{4}$$ $$\csc^2x=1+\Big(-\frac{4}{5}\Big)^2=1+\frac{16}{25}=\frac{41}{25}$$ $$\csc x=\pm\frac{\sqrt{41}}{5}$$ Since $\sec x\lt0$, $$\sec x=-\frac{\sqrt{41}}{4}$$ Since $\csc x\gt0$, $$\csc x=\frac{\sqrt{41}}{5}$$
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