Q:

Show that $\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$

I have to show that the left equation simplifies to $\tan\theta$:

Show that:
  $$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$ 

I do have prior knowledge that:
$$\tan \theta = \frac{\sin\theta}{\cos \theta}$$
But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far. 

A:

Using the double angle formulae:
$$\cos(2\theta)=1-2\sin^2\theta$$
$$\sin(2\theta)=2\sin\theta\cos\theta$$
$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

