QPaperGen

We use the identity:
\[
\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}
\]

Now, consider the left-hand side (LHS):
\begin{align*}
\tan 4x &= \tan(2 \cdot 2x) \\
&= \frac{2 \tan 2x}{1 - \tan^2 2x} \\
\text{Using } \tan 2x = \frac{2 \tan x}{1 - \tan^2 x}, \text{ we get:} \\
\tan 4x &= \frac{2 \cdot \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{4 \tan x}{1 - \tan^2 x} \cdot \frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2 - 4 \tan^2 x} \\
&= \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 - 2 \tan^2 x + \tan^4 x - 4 \tan^2 x} \\
&= \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x} \qquad
\textbf{(Hence, proved.)}
\end{align*}