Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 202: 79


There are 2 possible values for $$\frac{\sec x-\tan x}{\sin x}$$, which is $$\frac{\sec x-\tan x}{\sin x}=\frac{25\sqrt6-60}{12}$$ or $$\frac{\sec x-\tan x}{\sin x}=\frac{-25\sqrt6-60}{12}$$

Work Step by Step

$$\cos x=\frac{1}{5}\hspace{2cm}A=\frac{\sec x-\tan x}{\sin x}$$ To find values of $A$, we need to represent $A$ first in terms of $\cos x$ as much as we can. - Reciprocal Identity: $$\sec x=\frac{1}{\cos x}$$ - Quotient Identity: $$\tan x=\frac{\sin x}{\cos x}$$ Therefore, $$A=\frac{\frac{1}{\cos x}-\frac{\sin x}{\cos x}}{\sin x}$$ $$A=\frac{\frac{1-\sin x}{\cos x}}{\sin x}$$ $$A=\frac{1-\sin x}{\sin x\cos x}$$ So we have represented $A$ into $\sin x$ and $\cos x$. And we can find $\sin x$ according to $\cos x$. - Pythagorean Identity: $$\sin^2 x=1-\cos^2 x=1-(\frac{1}{5})^2=1-\frac{1}{25}=\frac{24}{25}$$ $$\sin x=\pm\frac{\sqrt{24}}{5}=\pm\frac{2\sqrt6}{5}$$ 1) As $\sin x=\frac{2\sqrt6}{5}$ and $\cos x=\frac{1}{5}$ $$A=\frac{1-\frac{2\sqrt6}{5}}{\frac{2\sqrt6}{5}\times\frac{1}{5}}$$ $$A=\frac{\frac{5-2\sqrt6}{5}}{\frac{2\sqrt6}{25}}$$ $$A=\frac{25(5-2\sqrt6)}{5\times2\sqrt6}$$ $$A=\frac{5(5-2\sqrt6)}{2\sqrt6}$$ $$A=\frac{5\sqrt6(5-2\sqrt6)}{2\times6}$$ (multiply both numerator and denominator by $\sqrt6$) $$A=\frac{25\sqrt6-60}{12}$$ 2) As $\sin x=-\frac{2\sqrt6}{5}$ and $\cos x=\frac{1}{5}$ $$A=\frac{1-(-\frac{2\sqrt6}{5})}{(-\frac{2\sqrt6}{5})\times\frac{1}{5}}$$ $$A=\frac{1+\frac{2\sqrt6}{5}}{-\frac{2\sqrt6}{25}}$$ $$A=-\frac{\frac{5+2\sqrt6}{5}}{\frac{2\sqrt6}{25}}$$ $$A=-\frac{25(5+2\sqrt6)}{5\times2\sqrt6}$$ $$A=-\frac{5(5+2\sqrt6)}{2\sqrt6}$$ $$A=-\frac{5\sqrt6(5+2\sqrt6)}{2\times6}$$ (multiply both numerator and denominator by $\sqrt6$) $$A=\frac{-25\sqrt6-60}{12}$$
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