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, \[ \boxed{\frac{\cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right)} \]