Answer
$\frac{\sec x-1}{\sin x \sec x}=\tan \frac{x}{2}$
Work Step by Step
$\frac{\sec x-1}{\sin x \sec x}=\tan \frac{x}{2}$
Rewrite $\sec x$ as $\frac{1}{\cos x}$:
$\frac{\frac{1}{\cos x}-1}{\sin x* \frac{1}{\cos x}}=\tan \frac{x}{2}$
Multiply top and bottom by $\cos x$:
$\frac{(\frac{1}{\cos x}-1)\cos x}{(\sin x* \frac{1}{\cos x}) \cos x}=\tan \frac{x}{2}$
Simplify:
$\frac{1-\cos x}{\sin x}=\tan \frac{x}{2}$
Use the half-angle formula for tangent on the bottom of page 556 and the identity will be proved:
$\tan \frac{x}{2}=\tan \frac{x}{2}$
