Explanation

(d)
Given, $\frac{\sin x}{\sin y} = \frac{1}{2},\frac{\cos x}{\cos y} = \frac{3}{2}$
$\Rightarrow \frac{\tan x}{\tan y} = \frac{1}{3}$
$\tan{(x + y) = \frac{\tan{x + \tan y}}{1 - \tan{x\tan y}} = \frac{4\tan x}{1 - 3\tan^{2}x}}$
Also, $\sin{y = 2\sin{x,\cos{y = \frac{2}{3}\cos x}}}$
$\Rightarrow \sin^{2}{y + \cos^{2}{y = 4\sin^{2}{x + \frac{4\cos^{2}x}{9} = 1}}}$
$\Rightarrow 36\tan^{2}{x + 4 = 9\sec^{2}{x = 9(1 + \tan^{2}{x)}}}$
$\Rightarrow 27{\ tan}^{2}{x = 5}$
$\Rightarrow \tan{x = \frac{\sqrt{5}}{3\sqrt{3}}}$
$\Rightarrow \tan{(x + y) = \frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1 - \frac{15}{27}} = \sqrt{15}}$