Prove that $\tan 4x = \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x}$.

We know that $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$.
$L.H.S. = \tan 4x = \tan 2(2x)$.
Using the formula $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$ with $A = 2x$:
$= \frac{2 \tan 2x}{1 - \tan^2 2x}$.
Substituting $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$:
$= \frac{2 \left( \frac{2 \tan x}{1 - \tan^2 x} \right)}{1 - \left( \frac{2 \tan x}{1 - \tan^2 x} \right)^2}$.
$= \frac{\frac{4 \tan x}{1 - \tan^2 x}}{1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}}$.
$= \frac{\frac{4 \tan x}{1 - \tan^2 x}}{\frac{(1 - \tan^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}}$.
$= \frac{4 \tan x (1 - \tan^2 x)}{(1 - \tan^2 x)^2 - 4 \tan^2 x}$.
$= \frac{4 \tan x (1 - \tan^2 x)}{1 + \tan^4 x - 2 \tan^2 x - 4 \tan^2 x}$.
$= \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x} = R.H.S.$