1. Fórmulas base:
$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$

2. Desarrollo:
Sea $x = 2\theta$:
$$ \tan 4\theta = \frac{2 \tan 2\theta}{1 - \tan^2 2\theta} $$
Sustituimos $\tan 2\theta$ por su equivalente en $\tan \theta$:
$$ \tan 4\theta = \frac{2 \left( \frac{2\tan\theta}{1-\tan^2\theta} \right)}{1 - \left( \frac{2\tan\theta}{1-\tan^2\theta} \right)^2} $$
Simplificamos el denominador:
$$ 1 - \frac{4\tan^2\theta}{(1-\tan^2\theta)^2} = \frac{(1-\tan^2\theta)^2 - 4\tan^2\theta}{(1-\tan^2\theta)^2} = \frac{1 - 2\tan^2\theta + \tan^4\theta - 4\tan^2\theta}{(1-\tan^2\theta)^2} $$
$$ = \frac{1 - 6\tan^2\theta + \tan^4\theta}{(1-\tan^2\theta)^2} $$
Sustituimos en la fracción principal:
$$ \tan 4\theta = \frac{\frac{4\tan\theta}{1-\tan^2\theta}}{\frac{1-6\tan^2\theta+\tan^4\theta}{(1-\tan^2\theta)^2}} = \frac{4\tan\theta(1-\tan^2\theta)}{1-6\tan^2\theta+\tan^4\theta} $$
$$ \tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta} $$

3. Conclusión:
Queda demostrada la identidad.
$$ \boxed{\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}} $$