Explanation

(b)
Let $\alpha,\ \beta$ are the roots of given equation $6x^{2} - 5x + 1 = 0$
$\Rightarrow \ \ \alpha + \beta = \frac{5}{6}$ and $\alpha\beta = \frac{1}{6}$
$\therefore\tan^{- 1}{\alpha + \tan^{- 1}{\beta = \tan^{- 1}\left( \frac{\alpha + \beta}{1 - \alpha\beta} \right)}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right) = \tan^{- 1}{(1) = \frac{\pi}{4}}$