Explanation

(b)
Given, $\left( \sqrt{3} - i \right) = (a + ib)(c + id)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (ac - bd) + i(ad + bc)$
On comparing the real and imaginary part on both sides, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ ac - bd = \sqrt{3}$
And $\ \ \ ad + bc = 1$
Now, $\tan^{- 1}{\left( \frac{b}{a} \right) + \tan^{- 1}\left( \frac{d}{c} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{bc + ad}{ac - bd} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{1}{\sqrt{3}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = n\pi + \frac{\pi}{6},\ n \in I$