How do you write $\sin(2x)$ in terms of $\tan x$?
Solution 1:
$$\begin{align}\sin2x&=2\sin x\cos x\\ &=\frac{2\sin x\cos x}{1}\\&=\frac{2\sin x\cos x}{\sin^2x+\cos^2x}\\&\small\text{(Divide both numerator and denominator by $\cos^2x$)}\\&=\frac{2\tan x}{\tan^2x+1}\qquad\small\left(\because\tan x=\frac{\sin x}{\cos x}\right)\end{align}$$