Prove that \[ \frac{\cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right) \]
Proof:
\[
LHS=\frac{\cos 2x}{1+\sin 2x}
\]
Using the identities:
\[
\cos 2x=\cos^2x-\sin^2x
\]
\[
\sin 2x=2\sin x\cos x
\]
Substituting these values:
\[
LHS=\frac{\cos^2x-\sin^2x}{1+2\sin x\cos x}
\]
Factorizing numerator:
\[
LHS=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}
\]
Cancel common factor:
\[
LHS=\frac{\cos x-\sin x}{\cos x+\sin x}
\]
Divide numerator and denominator by \(\cos x\):
\[
LHS=\frac{1-\tan x}{1+\tan x}
\]
Using the identity:
\[
\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\]
with
\[
A=\frac{\pi}{4}, \quad B=x
\]
we get
\[
\tan\left(\frac{\pi}{4}-x\right)
=\frac{1-\tan x}{1+\tan x}
\]
Therefore,
\[
LHS=\tan\left(\frac{\pi}{4}-x\right)
\]
Hence proved,
\[
\boxed{\frac{\cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right)}
\]