Explanation

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