Prove that:

$\mathrm{LHS}=\frac{\tan ^{2} 2 x-\tan ^{2} x}{1-\tan ^{2} 2 x \tan ^{2} x}$

$=\frac{(\tan 2 x+\tan x)(\tan 2 x-\tan x)}{1-\tan ^{2} 2 x \tan ^{2} x} \quad\left\{\right.$ Using $\left.A^{2}-B^{2}=(A+B)(A-B)\right\}$

$\tan 3 x=\tan (2 x+x)$ and $\tan x=\tan (2 x-x)$

$=\frac{\tan 3 x(1-\tan 2 x \tan x) \times \tan x(1+\tan 2 x \tan x)}{1-\tan ^{2} 2 x \tan ^{2} x}$

$[\because \tan 2 x+\tan x=\tan 3 x(1-\tan 2 x \tan x) \& \tan 2 x-\tan x=\tan x(1+\tan 2 x \tan x)]$

$=\frac{\tan 3 x \tan x\left(1-\tan ^{2} 2 x \tan ^{2} x\right)}{1-\tan ^{2} 2 x \tan ^{2} x}$

$=\tan 3 x \tan x$

= RHS

Hence proved.