# Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 694: 95

$\frac{2}{1+\cos x}-\tan^2 \frac{x}{2}=1$

#### Work Step by Step

Start with the left side: $\frac{2}{1+\cos x}-\tan^2 \frac{x}{2}$ Use the identity $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$: $=\frac{2}{1+\cos x}-\left(\frac{\sin x}{1+\cos x}\right)^2$ $=\frac{2}{1+\cos x}-\frac{\sin^2 x}{(1+\cos x)^2}$ Get a common denominator and subtract: $=\frac{2}{1+\cos x}*\frac{1+\cos x}{1+\cos x}-\frac{\sin^2 x}{(1+\cos x)^2}$ $=\frac{2(1+\cos x)}{(1+\cos x)^2}-\frac{\sin^2 x}{(1+\cos x)^2}$ $=\frac{2+2\cos x-\sin^2 x}{(1+\cos x)^2}$ Rewrite $\sin^2 x$ as $1-\cos^2 x$: $=\frac{2+2\cos x-(1-\cos^2 x)}{(1+\cos x)^2}$ Simplify: $=\frac{2+2\cos x-1+\cos^2 x}{(1+\cos x)^2}$ $=\frac{1+2\cos x+\cos^2 x}{(1+\cos x)^2}$ $=\frac{(1+\cos x)^2}{(1+\cos x)^2}$ $=1$ Since this equals the right side, the identity has been proven.

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