Explanation

(d)
$\tan^{- 1}{\left( \frac{\tan x}{4} \right) + \tan^{- 1}\left( \frac{3\sin{2x}}{5 + 3\cos{2x}} \right)}$
$= \tan^{- 1}{\left( \frac{\tan x}{4} \right) + \tan^{- 1}\left( \frac{\frac{6\tan x}{1 + \tan^{2}x}}{5 + \frac{3(1 - \tan^{2}x)}{1 + \tan^{2}x}} \right)}$
$= \tan^{- 1}{\left( \frac{\tan x}{4} \right) + \tan^{- 1}\left( \frac{6\tan x}{8 + 2\tan^{2}x} \right)}$
$= tan^{- 1}\ \left( \frac{\tan x}{4} \right) + \tan^{- 1}\left( \frac{3\tan x}{4 + \tan^{2}x} \right)$
$= \tan^{- 1}{\left( \frac{\frac{\tan x}{4} + \frac{3\tan x}{4 + \tan^{2}x}}{1 - \frac{3\tan^{2}x}{4(4 + \tan^{2}{x)}}} \right)\left( as\left| \frac{\tan x}{4} \bullet \frac{3\tan x}{4\tan^{2}x} \right| < 1 \right)}$
$= \tan^{- 1}\left( \frac{16\tan{x + \tan^{3}x}}{16 + \tan^{2}x} \right)$
$= \tan^{- 1}{(\tan{x) = x}}$