Answer
See below.
Work Step by Step
Given $z=tan\frac{\alpha}{2}$, we have $\frac{2z}{1+z^2}=\frac{2(tan\frac{\alpha}{2})}{1+(tan\frac{\alpha}{2})^2}=\frac{2(tan\frac{\alpha}{2})}{sec^2\frac{\alpha}{2}}=2tan\frac{\alpha}{2} cos^2\frac{\alpha}{2}=2sin\frac{\alpha}{2} cos\frac{\alpha}{2}=sin\alpha$
