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

You can try. 
$$\tan (3x) = \frac{\sin 3x}{\cos 3x} = \frac{\sin x \cos 2x + \cos x \sin 2x}{\cos x \cos 2x - \sin x\sin 2x} = \frac{\sin x(\cos^2 x- \sin^2x) + 2\sin x\cos^2 x}{\cos x(2\cos^2 x -1) - 2\sin^2 x\cos x} = \frac{\tan x(1-\tan^2 x) + 2\tan x}{2 - \frac{1}{\cos^2 x} - 2\tan^2 x} = \frac{3\tan x - \tan^3x}{1-3\tan^2x}$$
