Ask Me Help Desk - Hardest identity problem EVER!

The display is quite strange, I will interpret as
$\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}$
(if you meant another expression, repost using "math" environment)

The problem is lot easier than what seems, just try to simplify as much as you can before going with pure computations. Otherwise you get an unmanageable expression.

Use first $2\sin x\cos x=\sin 2x$

$\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}=\frac{1-(\sin 2x+\cos 2x)^2}{2\sin 2x\cos 2x}$

Numerator:

$1-(\sin 2x+\cos 2x)^2=1-(\sin^2 2x+\cos^2 2x +2\sin 2x\cos 2x)$
and using $\sin^2 2x+\cos^2 2x=1$ ,
$1-(\sin^2 2x+\cos^2 2x +2\sin 2x\cos 2x) =-2\sin 2x\cos 2x$

Denominator is $2\sin 2x\cos 2x$ , thus the original expression

$\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}=-1$