How do I simplify $\frac{\sin(2x)}{1-\cos (2x)}$? $$\dfrac{\sin(2x)}{1-\cos (2x)}$$
How do I simplify given expression?
My attempt: 
We know the double-angle identity for $\sin(2x)$ and $\cos(2x)$ as shown below
$$\sin(2x) = 2\sin (x)\cos(x)$$
$$\cos(2x) = 2\cos(x)-1$$
So we have that
$$\dfrac{2\sin (x)\cos(x)}{1-2\cos(x)-1} = \dfrac{2\sin(x)\cos(x)}{-2\cos^2(x)} = \dfrac{2\sin(x)}{-2\cos(x)} = -\tan (x)$$
I believe I've gone wrong somewhere.

We have
$$\cos(2x) = \cos^2x-\sin^2x=2\cos^\color{red}2(x)-1=1-2\sin^2 x$$
and then
$$\dfrac{\sin(2x)}{1-\cos (2x)}=\dfrac{2\sin x \cos x}{1-1+2\sin^2 x}=\cot x$$
