## Precalculus (6th Edition)

$\cos 2A=\frac{\cot A-\tan A}{\csc A\sec A}$
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.