# 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: $$\tan \theta = \frac{\cos \theta}{\sin \theta}$$

But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.

$$\cos(2\theta)=1-2\sin^2\theta$$ $$\sin(2\theta)=2\sin\theta\cos\theta$$
$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$
you need the double angle formulae $cos2\theta=2cos^2\theta-1$ and $sin2\theta=2sin\theta cos\theta$
use $cos2\theta$=$1-2{sin^2{\theta}}$ and $sin2\theta$=$2{sin\theta}{cos\theta}$ and you will get your required answer.