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) \[i\,\tan \frac{\theta }{2}\]

Correct Answer: C

Solution :

\[a=\cos \theta +i\sin \theta .\,\] \  \[\frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }.\,\] Rationalization of denominator, we get \[\frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\,\sin \theta }\times \frac{(1-\cos \theta )+i\sin \theta }{(1-\cos \theta )+i\sin \theta }\] \[=\frac{(1+\cos \theta )\,(1-\cos \theta )+(1+\cos \theta )\,i\sin \theta +(1-\cos \theta )i\sin \theta +{{i}^{2}}{{\sin }^{2}}\theta }{{{(1-\cos \theta )}^{2}}-{{(i\sin \theta )}^{2}}}\] \[=\frac{1-({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+2i\sin \theta }{1+({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )-2\,\cos \theta }\]\[=\frac{2i\sin \theta }{2(1-\cos \theta )}\] \[=\frac{i.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}}\]\[=i\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=i\cot \frac{\theta }{2}\].