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