A) \[\cot \theta \]
B) \[\cot \theta /2\]
C) \[i\cot \theta /2\]
D) \[i\tan \theta /2\]
Correct Answer: C
Solution :
\[a=\cos \theta +i\sin \theta ,\frac{1+a}{1-a}\] \[=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }\] \[=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }\times \frac{(1-\cos \theta )+i\sin \theta }{(1-\cos \theta )+i\sin \theta }\] \[=\frac{\begin{align} & (1+\cos \theta )(1-\cos \theta )+i\sin \theta (1-\cos \theta ) \\ & +\,(1+\cos \theta )i\sin \theta +{{i}^{2}}{{\sin }^{2}}\theta \\ \end{align}}{{{(1-\cos \theta )}^{2}}-{{(i\sin \theta )}^{2}}}\] \[=\frac{\begin{align} & 1-{{\cos }^{2}}\theta +i\sin \theta -i\sin \theta \cos \theta +i\sin \theta \\ & +i\sin \theta \cos \theta -{{\sin }^{2}}\theta \\ \end{align}}{1+{{\cos }^{2}}\theta -2\cos \theta +{{\sin }^{2}}\theta }\] \[=\frac{2i\sin \,\theta }{2(1-cos\theta )}\] \[=\frac{i2\sin \theta /2\cos \theta /2}{1-(1-2si{{n}^{2}}\theta /2)}=i\cot \,\theta /2\]
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