Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 54

$$\tan (2 x) =\frac{2\tan x}{1-\tan^2x}$$

We use the know identities to obtain: \begin{align*} \tan (2 x)&=\frac{\sin (2 x)}{\cos (2 x)}\\ &=\frac{2 \sin (x) \cos (x)}{\cos ^{2}(x)-\sin ^{2}(x)}\\ &= \frac{\frac{2 \sin (x) \cos (x)}{\cos^2 x}}{1-\frac{\sin ^{2}(x)}{\cos^2 x}}\\ &=\frac{2\tan x}{1-\tan^2x} \end{align*}