Explanation

(a)
$\therefore\cot^{- 1}{x + \sin^{- 1}{\frac{1}{\sqrt{5}} = \frac{\pi}{4}}}$
$\Rightarrow \tan^{- 1}{\frac{1}{x} + \tan^{- 1}{\frac{1}{2} = \tan^{- 1}1}}$
$\Rightarrow \tan^{- 1}{\frac{1}{x} = \tan^{- 1}{1 - \tan^{- 1}\frac{1}{2}}}$
$\Rightarrow \tan^{- 1}{\frac{1}{x} = \tan^{- 1}\left( \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} \right)}$
$\Rightarrow \tan^{- 1}{\frac{1}{x} = \tan^{- 1}\frac{1}{3}}$
$\Rightarrow x = 3$