• # question_answer The real part of  ${{(1-\cos \theta +2i\sin \theta )}^{-1}}$is [IIT 1978, 86] A) $\frac{1}{3+5\cos \theta }$ B) $\frac{1}{5-3\cos \theta }$ C) $\frac{1}{3-5\cos \theta }$ D) $\frac{1}{5+3\cos \theta }$

${{\{(1-\cos \theta )+i.2\sin \theta \}}^{-1}}={{\left\{ 2{{\sin }^{2}}\frac{\theta }{2}+i.4\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}}^{-1}}$ = ${{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}{{\left\{ \sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2} \right\}}^{-1}}$ $={{\left( 2\sin \frac{\theta }{2} \right)}^{-1}}\frac{1}{\sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2}}\times \frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}$ $=\frac{\sin \frac{\theta }{2}-i.2\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}\left( {{\sin }^{2}}\frac{\theta }{2}+4{{\cos }^{2}}\frac{\theta }{2} \right)}$. 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 }$