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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...

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If you are going to post to this site regularly, I suggest that the place to go is the faq, where you will find links to information on formatting mathematics on this site. – Gerry Myerson Nov 8 '12 at 2:41
Yes I'm definitely going to go through it ASAP as it makes it much easier to read.. It's just due to time today. – DavidSalib Nov 8 '12 at 2:44
@DavidSalib Kindly look here (…) on how to typeset your questions so that it is easier for people to read. – user17762 Nov 8 '12 at 3:15
up vote 4 down vote accepted

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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