A) 1/4
B) 1/2
C) tan q/2
D) 1/1- cos q
Correct Answer: B
Solution :
\[\frac{1}{1-\cos \,\,\theta +i\sin \,\,\theta }\]\[=\frac{1}{(1-\cos \,\theta )+i\sin \,\theta }\times \frac{(1-\cos \,\theta )-i\sin \,\theta }{(1-\cos \,\theta )-i\sin \,\theta }\] \[=\frac{(1-\cos \theta )-i\sin \theta }{{{(1-\cos \theta )}^{2}}+{{\sin }^{2}}\theta }\] \[=\frac{(1-\cos \theta )-i\sin \theta }{2(1-\cos \theta )}\] \[=\frac{(1-\cos \theta )}{2(1-\cos \theta )}-i\,\frac{\sin \theta }{2(1-\cos \theta )}.\] Therefore its real part = \[\frac{1-\cos \theta }{2\,(1-\cos \theta )}=\frac{1}{2}\]
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