• # question_answer If $a=\cos \,\theta +i\,\sin \,\theta ,$ then $\frac{1+a}{1-a}=$ [Karnataka CET 2000] A) $\cot \theta$ B) $\cot \frac{\theta }{2}$ C) $i\,\cot \frac{\theta }{2}$ D) $i\,\tan \frac{\theta }{2}$

$a=\cos \theta +i\sin \theta .\,$ \  $\frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }.\,$ 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 +{{\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 \frac{\theta }{2}}{\sin \frac{\theta }{2}}=i\cot \frac{\theta }{2}$.