<s>[INST] Trig Identities : $\frac{\sin (4x)}{1-\cos(4x)} \frac{1-\cos(2x)}{\cos(2x)} = \tan(x)$ 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...
 [/INST] Recall the following identities:
$$\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$$
$$1-\cos(2 \phi) = 2 \sin^2(\phi)$$
Make use of the above identities and you will get your solution.
Move your cursor over the gray area for complete solution.

\begin{align}\dfrac{\sin(4x)}{1-\cos(4x)} \dfrac{1 - \cos(2x)}{\cos(2x)} & = \dfrac{\sin(2(2x))}{1-\cos(2(2x))} \dfrac{1 - \cos(2x)}{\cos(2x)}\\ & = \dfrac{2 \sin(2x) \cos(2x)}{2 \sin^2(2x)} \dfrac{1-\cos(2x)}{\cos(2x)}\\ & = \dfrac{1-\cos(2x)}{\sin(2x)} ( \because \text{Cancelling } 2\sin(2x) \cos(2x))\\ & = \dfrac{2 \sin^2(x)}{2 \sin(x) \cos(x)}\\ & = \dfrac{\sin(x)}{\cos(x)} ( \because \text{Cancelling } 2\sin(x))\\ & = \tan(x) \end{align}

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