Explanation

(d)
$\left\{ \left( 1 - \cos\theta \right) + i.2\sin\theta \right\}^{- 1}$
$= \left( 2\sin^{2}{\frac{\theta}{2} + i.4\sin{\frac{\theta}{2}\cos\frac{\theta}{2}}} \right)^{- 1}$
$= \left( 2\sin\frac{\theta}{2} \right)^{- 1}\left( \sin{\frac{\theta}{2} + i2\cos\frac{\theta}{2}} \right)^{- 1}$
$= \left( 2\sin\frac{\theta}{2} \right)\frac{1}{\sin{\frac{\theta}{2} + i\ 2\cos\frac{\theta}{2}}} \times \frac{\sin{\frac{\theta}{2} - i2\cos\frac{\theta}{2}}}{\sin{\theta - i\ 2\cos\frac{\theta}{2}}}$
$= \frac{\sin{\frac{\theta}{2} - 2}i\cos\frac{\theta}{2}\ }{2\sin{\frac{\theta}{2}\ \left( \sin^{2}{\frac{\theta}{2} + 4\cos^{2}\frac{\theta}{2}} \right)}}$
$= \frac{\sin{\frac{\theta}{2} - 2}i\cos\frac{\theta}{2}}{2\sin{\frac{\theta}{2}\ \left( 1 + 3\cos^{2}\frac{\theta}{2} \right)}}$
$\therefore$ It’s real part is
$\frac{\sin\frac{\theta}{2}}{2\sin{\frac{\theta}{2}\left( 1 + 3\cos^{2}\frac{\theta}{2} \right)}} = \frac{1}{2\left\{ 1 + 3\left( \frac{\cos{\theta + 1}}{2} \right) \right\}}$
$= \frac{1}{5 + 3\cos\theta}$