Answer
$2\cos A-\sec A=\cos A-\frac{\tan A}{\csc A}$
Work Step by Step
Simplify the left side:
$2\cos A-\sec A$
$=2\cos A-\frac{1}{\cos A}$
$=\frac{2\cos^2 A}{\cos A}-\frac{1}{\cos A}$
$=\frac{2\cos^2 A-1}{\cos A}$
$=\frac{\cos 2A}{\cos A}$
Simplify the right side:
$\cos A-\frac{\tan A}{\csc A}$
$=\cos A-\frac{\frac{\sin A}{\cos A}}{\frac{1}{\sin A}}$
$=\cos A-\frac{\sin^2 A}{\cos A}$
$=\frac{\cos^2 A}{\cos A}-\frac{\sin^2 A}{\cos A}$
$=\frac{\cos^2 A-\sin^2 A}{\cos A}$
$=\frac{\cos 2A}{\cos A}$
Since both the left side and the right side are equal to $\frac{\cos 2A}{\cos A}$, they are equal to each other, and the identity is proven.