Explanation

(c)
Here, $\sqrt{1 - c^{2}} = nc - 1$
$\Rightarrow \ \ 1 - c^{2} = n^{2}c^{2} - 2nc + 1$
$\therefore\ \frac{c}{2n} = \frac{1}{1 + n^{2}}$ …(i)
or $\frac{c}{2n}\left( 1 + n\mathcal{z} \right)\left( 1 + \frac{n}{\mathcal{z}} \right) = \frac{1}{1 + n^{2}}\ \left\{ 1 + n^{2} + n\left( z + \frac{1}{\mathcal{z}} \right) \right\}$
$= \ \frac{1}{1 + n^{2}}\ \left\{ 1 + n^{2} + n\left( 2\cos\theta \right) \right\}$
$= \ \frac{\left( 1 + n^{2} \right) + 2n\cos\theta}{1 + n^{2}}$
$= 1 + \left( \frac{2n}{1 + n^{2}} \right)\cos\theta\ \lbrack$using Eq.(i)]
$= 1 + c\cos\theta$