Trigonometry- Double Angle Trigonometric Equations $$\tan(3x)=((\tan x(3-\tan^2x))/(1-3\tan^2x))$$ I'm just having a hard time seeing where the double angle formula fits in with verifying this equation.
Using a Adicional formula: $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y},$$ we have:

$$\tan(2x+x)=\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}=\frac{\tan (x+x)+\tan x}{1-\tan (x+x) \tan x}=\frac{\frac{\tan x + \tan x}{1-\tan x \tan x } +\tan x}{1-\frac{\tan x + \tan x}{1-\tan x \tan x } \tan x}$$
$$=\frac{\tan x +\frac{2\tan x}{1-\tan^2 x}}{1-\tan x \frac{2\tan x}{1-\tan^2 x}}=\frac{\frac{\tan x -\tan^3 x + 2\tan x}{1-\tan^2 x}}{\frac{1-\tan^2 x-2\tan^2 x}{1-\tan^2 x}}$$
$$=\frac{{\tan x -\tan^3 x + 2\tan x}}{{1-\tan^2 x-2\tan^2 x}}=\frac{{3\tan x -\tan^3 x }}{{1-3\tan^2 x}}=\frac{\tan x (3-\tan^2 x)}{1-3\tan^2 x}$$
$$\tan(3x)=\frac{\sin(2x+x)}{\cos(2x+x)}=\frac{\sin(2x)\cos x + \sin x \cos(2x)}{\cos(2x)\cos x -\sin(2x)\sin x}=\frac{3 \sin x -4 \sin ^3 x}{-3 \cos x + 4 \cos ^3 x}=(\tan x) \frac{3 -4 \sin ^2 x}{-3 +4 \cos ^2 x}$$