Explanation

(d)
$4\tan^{- 1}{\frac{1}{5} = 2\left\lbrack 2\tan^{- 1}\ \frac{1}{5} \right\rbrack}$
$= 2\tan^{- 1}{\frac{\frac{2}{5}}{1 - \frac{1}{25}} = 2\tan^{- 1}\frac{5}{12}}$
$= \tan^{- 1}\frac{\frac{10}{12}}{1 - \frac{25}{144}}$
$= \tan^{- 1}\frac{120}{119}$
So, 4 $\tan^{- 1}{\frac{1}{5} - \tan^{- 1}{\frac{1}{239} = \tan^{- 1}{\frac{120}{119} - \tan^{- 1}\frac{1}{239}}}}$
$= \tan^{- 1}\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}}$
$= \tan^{- 1}\frac{(120 \times 239) - 119}{(119 \times 239) + 120}$
$= \tan^{- 1}{\frac{28561}{28561} = \tan^{- 1}{1 = \frac{\pi}{4}}}$