Explanation

(d)
Given, $\tan{x = \frac{b}{a}}$
$\therefore{acos}{2x = b\sin{2x = a \times \frac{1 - \tan^{2}x}{1 + \tan^{2}x} + b \times \frac{2\tan x}{1 + \tan^{2}x}}}$
$= a \times \frac{1 - \frac{b^{2}}{a^{2}}}{1 + \frac{b^{2}}{a^{2}}} + b \times \frac{2\frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}}$
$= \frac{a(a^{2} - b^{2})}{a^{2} + b^{2}} + \frac{2ab^{2}}{a^{2} + b^{2}}$
$= \frac{a(a^{2} + b^{2})}{a^{2} + b^{2}} = a$