Explanation

(b)
We have,
$T_{4} = \ ^{5}C_{3}\left( \frac{1}{x} \right)^{5 - 3}\left( x\tan x^{3} \right) = 10x\tan^{3}x$
and $T_{2} = \ ^{5}C_{2}\left( \frac{1}{x} \right)^{5 - 1}\left( x\tan x \right) = \frac{5\ \tan x}{x^{3}}$
Given $\frac{T_{4}}{T_{2}} = \frac{2}{27}\pi^{4}\ \Rightarrow \ 2x^{4}\tan^{2}x = \frac{2}{27}\pi^{4}$
$\Rightarrow \ x^{2}\tan x = \pm \frac{1}{3\sqrt{3}}\pi^{2}$
It is possible (from among the answers) when $x = \pm \frac{\pi}{3}$