Prove that

To Prove: $\frac{\tan 2 x}{1+\sec 2 x}=\tan x$

Taking LHS,

$=\frac{\frac{\sin 2 x}{\cos 2 x}}{1+\frac{1}{\cos 2 x}}\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta} \& \sec \theta=\frac{1}{\cos \theta}\right]$

$=\frac{\sin 2 x}{\cos 2 x\left(\frac{\cos 2 x+1}{\cos 2 x}\right)}$

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

$=\frac{2 \sin x \cos x}{1+\cos 2 x}[\because \sin 2 x=2 \sin x \cos x]$

$=\frac{2 \sin x \cos x}{2 \cos ^{2} x}\left[\because 1+\cos 2 x=2 \cos ^{2} x\right]$

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

$=\tan \times\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

= RHS

∴ LHS = RHS

Hence Proved