Explanation

(c)
Since, $a = \cos{\theta + i\sin\theta}$
$\therefore\ \ \ \frac{1 + a}{1 - a} = \frac{1 + \cos{\theta + i\sin\theta}}{1 - \cos{\theta - i\sin\theta}}$
$= \frac{\left\lbrack \left( 1 + \cos\theta \right) + i\sin\theta \right\rbrack\lbrack\left( 1 - \cos\theta \right) + i\sin\theta\rbrack}{\left\lbrack \left( 1 - \cos\theta \right) - i\sin\theta \right\rbrack\lbrack\left( 1 - \cos\theta \right) + i\sin\theta\rbrack}$
$= \frac{2i\sin\theta}{\left( 1 - \cos\theta \right)^{2} + \sin^{2}\theta}$
$= \frac{i.4\sin{\frac{\theta}{2}\cos\frac{\theta}{2}}}{4\sin^{2}\frac{\theta}{2}} = i\cot\frac{\theta}{2}$