A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\tan \theta /2\]
D) \[\frac{1}{1-\cos \theta }\]
Correct Answer: B
Solution :
\[z=\frac{1}{1-\cos \theta +i\sin \theta }\] \[\Rightarrow \]\[z=\frac{1}{(1-\cos \theta )+i\sin \theta }\] \[\times \frac{\,(1-\cos \theta )-i\sin \theta }{(1-\cos \theta )-i\sin \theta }\] \[\Rightarrow \]\[z=\frac{(1-\cos \theta )-i\sin \theta }{{{(1-\cos \theta )}^{2}}-{{i}^{2}}{{\sin }^{2}}\theta }\] \[\Rightarrow \]\[z=\frac{(1-cos\theta )-isin\theta }{1+{{\cos }^{2}}\theta -2\cos \theta +si{{n}^{2}}\theta }\] \[=\frac{(1-\cos \theta )-i\sin \theta }{2(1-\cos \theta )}\] \[=\frac{1-\cos \theta }{2(1-\cos \theta )}-\frac{i\sin \theta }{2(1-\cos \theta )}\] The real part \[=1/2\]
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