Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way. 
I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
And I am stuck...

$\sin(2x) = 2\sin(x)\cos(x)$
$\sin(2x) =\dfrac{2\sin(x)\cos(x)}{\sin^2(x)+\cos^2(x)}$
divide throughout by $\cos^2(x)$
$\sin(2x) = \dfrac{2\tan(x)}{1+\tan^2(x)}$
let $x\to\frac x2$ to get the desired answer
