#### Answer

$2\cos^3 x-\cos x=\frac{\cos^2 x-\sin^2 x}{\sec x}$

#### Work Step by Step

Start with the right side:
$\frac{\cos^2 x-\sin^2 x}{\sec x}$
Rewrite in terms of sine and cosine:
$=\frac{\cos^2 x-\sin^2 x}{\frac{1}{\cos x}}$
Multiply top and bottom by $\cos x$:
$=\frac{\cos^2 x-\sin^2 x}{\frac{1}{\cos x}}*\frac{\cos x}{\cos x}$
$=\cos x*(\cos^2 x-\sin^2 x)$
Rewrite $\sin^2 x$ as $1-\cos^2 x$:
$=\cos x*(\cos^2 x-(1-\cos^2 x))$
Simplify:
$=\cos x*(\cos^2 x-1+\cos^2 x)$
$=\cos x*(2\cos^2 x-1)$
$=2\cos^3 x-\cos x$
Since this equals the left side, the identity has been proven.