Answer
$\sec 2x=\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}$
Work Step by Step
Start with the right side:
$\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}$
Multiply top and bottom by $\cos^4 x$:
$=\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}*\frac{\cos^4 x}{\cos^4 x}$
$\frac{\sec^2 x\cos^4 x+\sec^4 x\cos^4 x}{2\cos^4 x+\sec^2 x\cos^4 x-\sec^4 x\cos^4 x}$
$=\frac{\cos^2 x+1}{2\cos^4 x+\cos^2 x-1}$
Factor the denominator and simplify:
$=\frac{\cos^2 x+1}{(2\cos^2 x-1)(\cos^2 x+1)}$
$=\frac{1}{2\cos^2 x-1}$
Use the identity $2\cos^2 x-1=\cos 2x$:
$=\frac{1}{\cos 2x}$
$=\sec 2x$
Since this equals the left side, the identity has been proven.
