# Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 68

$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$

#### Work Step by Step

Start with the left side: $\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}$ Factor out $\frac{1}{2}$: $=\frac{1}{2}(\cot\frac{x}{2}-\tan\frac{x}{2})$ Write cotangent as 1 divided by tangent: $=\frac{1}{2}(\frac{1}{\tan\frac{x}{2}}-\tan\frac{x}{2})$ Use half-angle identities for tangent: $=\frac{1}{2}\left(\frac{1}{\frac{\sin x}{1+\cos x}}-\frac{1-\cos x}{\sin x}\right)$ Simplify: $=\frac{1}{2}(\frac{1+\cos x}{\sin x}-\frac{1-\cos x}{\sin x})$ $=\frac{1}{2}*\frac{(1+\cos x)-(1-\cos x)}{\sin x}$ $=\frac{1}{2}*\frac{1+\cos x-1+\cos x}{\sin x}$ $=\frac{1}{2}*\frac{2\cos x}{\sin x}$ $=\frac{\cos x}{\sin x}$ $=\cot x$ Since this equals the right side, the identity has been proven.

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