Here we will show an expansion for \(\tan(2x)\).
Note that for any angle:
$$\tan x = \frac{\sin x}{\cos x}$$
Therefore
$$\tan 2x = \frac{\sin 2x}{\cos 2x}$$
We can use the double-angle identities for sine and cosine to rewrite this as
$$\tan 2x = \frac{2\sin x\cos x}{1 - 2\sin^2 x}$$
Now, divide both the numerator and the denominator of the fraction by \(\cos^2 x\), which gives
$$\tan 2x = \frac{\frac{2\sin x\cos x}{\cos^2 x}}{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{2\tan x}{1 - \tan^2 x}$$
This is the end of the proof. We see that
$$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$$