Answer
$\tan (\frac{x}{2})+\cos x\tan (\frac{x}{2})=\sin x$
Work Step by Step
Start with the left side:
$\tan (\frac{x}{2})+\cos x\tan (\frac{x}{2})$
Use the identity $\tan (\frac{x}{2})=\frac{1-\cos x}{\sin x}$:
$=\frac{1-\cos x}{\sin x}+\cos x*\frac{1-\cos x}{\sin x}$
Simplify:
$=\frac{1-\cos x}{\sin x}+\frac{\cos x(1-\cos x)}{\sin x}$
$=\frac{1-\cos x}{\sin x}+\frac{\cos x-\cos^2 x}{\sin x}$
Add the two fractions:
$=\frac{1-\cos x+\cos x-\cos^2 x}{\sin x}$
Simplify:
$=\frac{1-\cos^2 x}{\sin x}$
Use the identity $1-\cos^2 x=\sin^2 x$:
$=\frac{\sin^2 x}{\sin x}$
$=\sin x$
Since this equals the right side, the identity is proven.
