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

Solution:

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

$\Rightarrow \frac{\tan \mathrm{x}}{\tan \mathrm{y}}=\frac{1}{3}$

$\begin{array}{r}\tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{xtan}\mathrm{y}}=\frac{4\tan \mathrm{x}}{1-3{\tan }^2\mathrm{x}}\\ \text{ Also, }\sin \mathrm{y}=2\sin \mathrm{x},\cos \mathrm{y}=\frac{2}{3}\cos \mathrm{x}\end{array}$

$\begin{array}{l}\Rightarrow {\sin }^2\mathrm{y}+{\cos }^2\mathrm{y}=4{\sin }^2\mathrm{x}+\frac{4{\cos }^2\mathrm{x}}{9}=1\\ \Rightarrow 36{\tan }^2\mathrm{x}+4=9{\sec }^2\mathrm{x}=9\left(1+{\tan }^2\mathrm{x}\right)\\ \Rightarrow 27{\tan }^2\mathrm{x}=5\\ \Rightarrow \tan \mathrm{x}=\frac{\sqrt{5}}{3\sqrt{3}}\\ \Rightarrow \tan (\mathrm{x}+\mathrm{y})=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-\frac{15}{27}}=\sqrt{15}\end{array}$