Double angle Prove question

Prove that:
sin(2x)
---=cot(x)
1-cos(2x)

I've worked it down to simply cosx on the left side, but the other side is very annoying and confusing.
 
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Soroban

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May 2006
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Lexington, MA (USA)
Hello, tallguywhoplaysguitar!

We need some double-angle identities:

. . \(\displaystyle \sin2x \;=\;2\sin x\cos x\)

. . \(\displaystyle \sin^2\!x \:=\:\frac{1-\cos2x}{2} \quad\Rightarrow\quad 1 - \cos2x \:=\:2\sin^2\!x\)


Prove that: .\(\displaystyle \frac{\sin2x}{1-\cos2x} \:=\:{\color{red}\cot x}\)


We have: . \(\displaystyle \frac{\sin2x}{1-\cos2x}\;\;\begin{array}{c}\to \\ \to \end{array} \;\; \frac{2\sin x\cos x}{2\sin^2\!x} \;\;=\;\;\frac{\cos x}{\sin x} \;\;=\;\;\cot x \)

 
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