Verify the triple angle formula Verify the triple angle formula
$$\tan(3x) = \frac{3 \tan(x) − \tan^3(x)}{1 − 3 \tan^2(x)}$$
I have tried simplifying the right side by the following
$$\tan(3x) = \frac{\tan(x)(3 − \tan^2(x))}{1 − 3 \tan^2(x)}$$
but then I am getting stuck trying to verify the equation

If you are allowed to use complex number, there is a simple way to derive the
triple angle formula. For real $x$ with $\cos x, \cos 3x \ne 0$, if one expand $e^{i3x}$ in two different ways, one get
$$\cos 3x( 1 + i\tan 3x)
=\cos 3x + i\sin 3x\\
= e^{i3x} = (e^{ix})^3\\ = (\cos x + i\sin x)^3 = \cos^3x ( 1 + i\tan x)^3$$
If one compare the ratio of real and imaginary part of both sides, one immediately obtain
$$\tan x = \frac{\Im (1 + i\tan x)^3 }{\Re (1+i\tan x)^3} = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$
You can use the same trick to expand $\tan n x$ for other integer $n > 3$.
