If $\frac{\sin \mathrm{x}}{\sin \mathrm{y}}=\frac{1}{2},\frac{\cos \mathrm{x}}{\cos \mathrm{y}}=\frac{3}{2}\text{, where }\mathrm{x},\mathrm{y}\in \left(0,\frac{\mathrm{\pi }}{4}\right)\text{, }$then the value of tan(x + y) is equal to

$\frac{\sin \mathrm{x}}{\sin \mathrm{y}}=\frac{1}{2},\frac{\cos \mathrm{x}}{\cos \mathrm{y}}=\frac{3}{2}\Rightarrow \frac{\tan \mathrm{x}}{\tan \mathrm{y}}=\frac{1}{3}$

$\Rightarrow \tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{xtan}\mathrm{y}}=\frac{4\tan \mathrm{x}}{1-3{\tan }^2\mathrm{x}}$
Also $\sin \mathrm{y}=2\sin \mathrm{x},\cos \mathrm{y}=\frac{2}{3}\cos \mathrm{x}$
or $36{\tan }^2\mathrm{x}+4=9{\sec }^2\mathrm{x}\text{ or }\tan \mathrm{x}=\frac{\sqrt{5}}{3\sqrt{3}}$
or $\tan (\mathrm{x}+\mathrm{y})=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-\frac{15}{27}}=\frac{4\sqrt{5}\times 27}{12\times 3\sqrt{3}}=\sqrt{15}$