I want to prove that $\frac{\sin (4x)}{1-\cos(4x)} \frac{1-\cos(2x)}{\cos(2x)} = \tan(x)$
\begin{align} \text{Left hand side} : & = \sin(2x+2x)/(1-\cos(2x+2x)) \times ((1-\cos^2x+\sin^2x)/(\cos^2x-\sin^2x))\\ & = ((2\sin^2x)(\cos^2x))/(2\sin^2(2x)) \times (2\sin^2x/(2\cos^2x -1))\\ & = 2\sin^2(x)\cos^2(x)/\sin^2(2x) \end{align}
Not sure where to go from here...