Answer
$$\sec2x=\frac{\sec^2x+\sec^4x}{2+\sec^2x-\sec^4x}$$
The equation is an identity, as proved below.
Work Step by Step
$$\sec2x=\frac{\sec^2x+\sec^4x}{2+\sec^2x-\sec^4x}$$
We tackle the more complicated side, which is the right side.
$$X=\frac{\sec^2x+\sec^4x}{2+\sec^2x-\sec^4x}$$
$\sec x$ can be written as $\sec x=\frac{1}{\cos x}$. Therefore,
$$\sec^2x=\frac{1}{\cos^2x}\hspace{2cm}\sec^4x=\frac{1}{\cos^4x}$$
$$X=\frac{\frac{1}{\cos^2x}+\frac{1}{\cos^4x}}{2+\frac{1}{\cos^2x}-\frac{1}{\cos^4x}}$$
$$X=\frac{\frac{\cos^2x+1}{\cos^4x}}{\frac{2\cos^4x+\cos^2x-1}{\cos^4x}}$$
$$X=\frac{(\cos^2x+1)\cos^4x}{(2\cos^4x+\cos^2x-1)\cos^4x}$$
$$X=\frac{\cos^2x+1}{2\cos^4x+\cos^2x-1}$$
We examine the denominator:
$$2\cos^4x+\cos^2x-1=2\cos^4x+2\cos^2x-\cos^2x-1$$
$$2\cos^4x+\cos^2x=2\cos^2x(\cos^2x+1)-(\cos^2x+1)$$
$$2\cos^4x+\cos^2x=(\cos^2x+1)(2\cos^2x-1)$$
Thus, $$X=\frac{\cos^2x+1}{(\cos^2x+1)(2\cos^2x-1)}$$
$$X=\frac{1}{2\cos^2x-1}$$
From Double-Angle Identity: $2\cos^2x-1=\cos2x$
$$X=\frac{1}{\cos2x}$$
From Reciprocal Identity: $\frac{1}{\cos2x}=\sec2x$
$$X=\sec2x$$
Therefore the equation has been verified to be an identity.
