Answer
$\cos 2A=\frac{\cot A-\tan A}{\csc A\sec A}$
Work Step by Step
Start with the right side:
$\frac{\cot A-\tan A}{\csc A\sec A}$
Rewrite in terms of sine and cosine:
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}}{\frac{1}{\sin A}*\frac{1}{\cos A}}$
Multiply top and bottom by $\sin A\cos A$:
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}}{\frac{1}{\sin A}*\frac{1}{\cos A}}*\frac{\sin A\cos A}{\sin A\cos A}$
Simplify:
$=\frac{\cos^2 A-\sin^2 A}{1}$
$=\cos^2 A-\sin^2 A$
$=\cos 2A$
Since this equals the left side, the identity has been proven.
