Problem 74 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

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 1: 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 \(1 - \sin^2x = \cos^2x \).

Step 2: 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 \(-2\sin x \cdot 2\sec^2 x\).

Step 3: 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 \(-2 \tan x \cdot \sec x \cdot 2\sec^2 x = -4 \tan x \sec^3 x\).