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Chapter 5: Problem 32

Verify the identity. $$\cos x-\frac{\cos x}{1-\tan x}=\frac{\sin x \cos x}{\sin x-\cos x}$$

Short Answer

Expert verified
The given expression is true. The simplified versions of the left and right side of the equation are the same.

Step by step solution

01

Simplify the Left Side of the Equation

We start with the left side of the equation and simplify the expression. Recall that \( \tan x = \frac{\sin x}{\cos x} \). So, we can rewrite the term \( \frac{\cos x}{1-\tan x} \) as \( \frac{\cos x}{1-\frac{\sin x}{\cos x}} \). To eliminate the fraction in the denominator, multiply both the numerator and the denominator by \( \cos x \), getting \( \frac{\cos^2 x}{\cos x - \sin x} \). Now the left side of the equation is \( \cos x - \frac{\cos^2 x}{\cos x - \sin x} \).
02

Simplify the Right Side of the Equation

Now, we simplify the right side of the equation, \( \frac{\sin x \cos x}{\sin x-\cos x} \), by multiplying the numerator and the denominator by \( -1 \) to remove the signs. This results in \( \frac{-\sin x \cos x}{\cos x - \sin x} \). The expressions for the right side and left side of the equation are identical, but opposite in sign.
03

Combine Left Side Terms

Now, we combine the terms on the left side to find a common denominator. \( \cos x = \frac{\cos x (\cos x - \sin x)}{\cos x - \sin x} \), which simplifies to \( \frac{\cos^2 x - \sin x \cos x}{\cos x - \sin x} \). This expression is the same as the right side when multiplying both numerator and denominator by -1. Hence, it's confirmed that \( \cos x-\frac{\cos x}{1-\tan x} = \frac{\sin x \cos x}{\sin x-\cos x} \).

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