Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 660: 74

Answer

The expression in terms of the function $\sec x\ \text{ and }\ \tan x$ is $-4\sec x\tan x$.

Work Step by Step

By multiplying the part of expression $\frac{1-\sin x}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ and the other part of expression $\frac{1+\sin x}{1-\sin x}$ by $\frac{1+\sin x}{1+\sin x}$: $\begin{align} & \frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x}=\frac{1-\sin x}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}-\frac{1+\sin x}{1-\sin x}\times \frac{1+\sin x}{1+\sin x} \\ & =\frac{1-\sin x-\sin x+{{\sin }^{2}}x}{1-{{\sin }^{2}}x}-\frac{1+\sin x+\sin x+{{\sin }^{2}}x}{1-{{\sin }^{2}}x} \\ & =\frac{1-2\sin x+{{\sin }^{2}}x-\left( 1+2\sin x+{{\sin }^{2}}x \right)}{1-{{\sin }^{2}}x} \\ & =\frac{1-2\sin x+{{\sin }^{2}}x-1-2\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x} \end{align}$ Further solve: $\frac{1-2\sin x+{{\sin }^{2}}x-1-2\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x}=\frac{-4\sin x}{1-{{\sin }^{2}}x}$ Now, apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , $1-si{{n}^{2}}x$ can be written as ${{\cos }^{2}}x$: $\begin{align} & \frac{-4\sin x}{1-{{\sin }^{2}}x}=\frac{-4\sin x}{{{\cos }^{2}}x} \\ & =-4\cdot \frac{1}{\cos x}\cdot \frac{\sin x}{\cos x} \end{align}$ By using the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$ , then the above expression can be further simplified as: $\begin{align} & -4\cdot \frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}=-4\sec x\cdot \tan x \\ & =-4\sec x\tan x \end{align}$
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