A) \[\frac{1}{3+5\cos \theta }\]
B) \[\frac{1}{5-3\cos \theta }\]
C) \[\frac{1}{3-5\cos \theta }\]
D) \[\frac{1}{5+3\cos \theta }\]
Correct Answer: D
Solution :
\[{{\{(1-\cos \theta )+i.2\sin \theta \}}^{-1}}={{\left\{ 2{{\sin }^{2}}\frac{\theta }{2}+i.4\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}}^{-1}}\] = \[{{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}{{\left\{ \sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2} \right\}}^{-1}}\] \[={{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}\frac{1}{\sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2}}\times \frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}\] \[=\frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}\left( {{\sin }^{2}}\frac{\theta }{2}+4{{\cos }^{2}}\frac{\theta }{2} \right)}\]. it?s real part \[=\frac{\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}\left( 1+3{{\cos }^{2}}\frac{\theta }{2} \right)}=\frac{1}{2\left( 1+3{{\cos }^{2}}\frac{\theta }{2} \right)}\]\[=\frac{1}{5+3\cos \theta }\]
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