Prove the identity: $\dfrac{\tan 2x \tan x}{\tan 2x - \tan x} = \sin 2x$
$\begin{aligned}
LHS & = \dfrac{\tan 2x \tan x}{\tan 2x - \tan x} \\\\
& = \dfrac{\dfrac{2 \tan x}{1 - \tan^2 x} \times \tan x}{\dfrac{2 \tan x}{1 - \tan^2 x} - \tan x} \\\\
& = \dfrac{2 \tan^2 x}{2 \tan x - \tan x + \tan^3 x} \\\\
& = \dfrac{2 \tan^2 x}{\tan x + \tan^3 x} \\\\
& = \dfrac{2 \tan^2 x}{\tan x \left(1 + \tan^2 x\right)} \\\\
& = \dfrac{2 \tan x}{\sec^2 x} \\\\
& = \dfrac{\dfrac{2 \sin x}{\cos x}}{\dfrac{1}{\cos^2 x}} \\\\
& = 2 \sin x \cos x \\\\
& = \sin 2x = RHS
\end{aligned}$
Hence proved.