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