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:
$$\cos(2\theta)=1-2\sin^2\theta$$
$$\sin(2\theta)=2\sin\theta\cos\theta$$
$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$
A: you need the double angle formulae 
$
cos2\theta=2cos^2\theta-1$ and $
sin2\theta=2sin\theta cos\theta 
$
A: use $cos2\theta$=$1-2{sin^2{\theta}}$ and $sin2\theta$=$2{sin\theta}{cos\theta}$ and you will get your required answer.
A: Here is a detailed answer.Let's go!
$$
\require{cancel}
\begin{align}
\frac{1-\cos2\theta}{\sin2\theta}&=\frac{1-\left(\cos^2\theta-\sin^2\theta\right)}{2\sin\theta\cos\theta}\\
&=\frac{1-\cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\sin^2\theta\cancel{+\cos^2\theta}\cancel{-\cos^2\theta}+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\bcancel2\cancelto{\sin\theta}{\sin^2\theta}}{\bcancel2\cancel{\sin\theta}\cos\theta}\\
&=\frac{\sin\theta}{\cos\theta}\\
&=\tan\theta
\end{align}
$$
I hope this helps.
