If $a=\cos \theta +i\sin \theta ,$ then $\frac{1+a}{1-a}=$

$a=\cos \theta +i\sin \theta .$

\  $\frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }.$

Rationalization of denominator, we get

$\frac{1+a}{1-a}=\frac{(1+\cos \theta )+i \sin \theta }{(1-\cos \theta )-i\sin \theta }\times \frac{(1-\cos \theta )+i \sin \theta }{(1-\cos \theta )+i \sin \theta }$

$=\frac{(1+\cos \theta )(1-\cos \theta )+(1+\cos \theta )i \sin \theta +(1-\cos \theta )i \sin \theta +i^2{\sin }^2\theta }{{(1-\cos \theta )}^2-{(i\sin \theta )}^2}$

$=\frac{1-{\cos }^2 \theta +i \sin \theta +i \sin \theta \cos \theta +i \sin \theta -i \sin \theta \cos \theta -{\sin }^2 \theta }{1+{\cos }^2 \theta -2\cos \theta +{\sin }^2 \theta }$

$=\frac{1-({\cos }^2\theta +{\sin }^2\theta )+2i \sin \theta }{1+({\cos }^2\theta +{\sin }^2\theta )-2\cos \theta }$$=\frac{2i \sin \theta }{2(1-\cos \theta )}$

$=\frac{i.2 \sin \frac{\theta }{2} \cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$$=i\frac{\cos {\displaystyle }\frac{\theta }{2}}{\sin {\displaystyle }\frac{\theta }{2}}=i \cot \frac{\theta }{2}.$