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Chapter 5: Problem 74

Rewrite each expression in terms of the given function or functions. $$\frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x} ; \sec x \text { and } \tan x$$

Short Answer

Expert verified
The given expression \( \frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x} \) can be rewritten in terms of secant and tangent functions as: \( -4 \tan x \sec^3 x \) .

Step by step solution

01

Simplify the given expression

First, subtract the two fractions by finding a common denominator and simplifying:\n\[ \frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x} = \frac{(1-\sin x)^2-(1+\sin x)^2}{(1+\sin x)(1-\sin x)} \]\n This simplifies to:\n\[ \frac{-4\sin x}{1-\sin^2x} \],\n which further simplifies to \n\[ -2\sin x \cdot \frac{2}{\cos^2x} .\] because \(1 - \sin^2x = \cos^2x \).
02

Translate in terms of secant function

\( \sec x \) is the reciprocal of \( \cos x \), so \( \sec^2 x \) is the reciprocal of \( \cos^2 x \). Therefore, \( \frac{2}{\cos^2x} \) becomes \( 2\sec^2 x \). So the expression simplifies to \(-2\sin x \cdot 2\sec^2 x\).
03

Translate in terms of tangent function

\( \tan x \) is \( \frac{\sin x}{\cos x} \). Therefore, \( \sin x \) is \( \tan x \cdot \cos x \). Given that \( \cos x \) is the reciprocal of \( \sec x \), \( \sin x \) becomes \( \tan x \cdot \sec x \). So the final expression is \(-2 \tan x \cdot \sec x \cdot 2\sec^2 x = -4 \tan x \sec^3 x\).

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Most popular questions from this chapter

Exercises \(116-118\) will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$\sin \pi \cos \frac{\pi}{2}=\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{2}\right)+\sin \left(\pi-\frac{\pi}{2}\right)\right]$$

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Will help you prepare for the material covered in the next section.$$\text { Give exact values for } \sin 30^{\circ}, \cos 30^{\circ}, \sin 60^{\circ}, \text { and } \cos 60^{\circ}$$
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