Prove that:

$\mathrm{LHS}=\frac{\cos 2 x}{1+\sin 2 x}$

$=\frac{\cos ^{2} x-\sin ^{2} x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \times \cos x} \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right.$ and $\left.\sin ^{2} x+\cos ^{2} x=1\right]$

$=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^{2}} \quad\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\frac{\cos x-\sin x}{\cos x+\sin x}$

On dividing the numerator and denominator by cos x, we get

$=\frac{1-\tan x}{1+\tan x}$

$=\tan \left(\frac{\pi}{4}-\mathrm{x}\right)=\mathrm{RHS}$

Hence proved.