If  $x,y\in (0,\frac{\pi }{2})$ and  $2\sin x=\sin y,2\cos x=3\cos y$  then  $\left[\tan (x+y)\right]$ (where)[.] denote the G.I.F) is 

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

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

${\sin }^{2}y+{\cos }^{2}y=4{\sin }^{2}x+4\frac{{\cos }^{2}x}{9}=1$

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

$$\begin{gathered} \tan (x+y)=\frac{4\tan x}{1-3{\tan }^{2}x} \\ {\tan }^2(x+y)=15,\tan (x+y)=\sqrt{15} \\ and \left[\tan (x+y)\right]=3 \end{gathered}$$