Find the value of $\frac{\tan A}{\tan B}$, given $\frac{\sin A}{\sin B}=5$ 
If $\displaystyle \frac{\sin A}{\sin B}=5$, then find the value of $\displaystyle \frac{\tan A}{\tan B}$

Try using the Componendo and Dividendo formula:
$$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3}{2}$$
$$\frac{\tan(A+B)/2}{\tan(A-B)/2}=\frac{3}{2}$$
Can someone help me find: $$\frac{\tan A}{\tan B}$$ 
Thanks

This is impossible to answer for general $A$ and $B$ satisfying the condition.
If $\displaystyle \frac{\tan A}{\tan B}=k$, then $\displaystyle k=\frac{\sin A\cos B}{\cos A\sin B}=\frac{5\cos B}{\cos A}$.
$$5\cos B=k\cos A$$
$$(5\sin B)^2+(5\cos B)^2=\sin^2A+k^2\cos^2A$$
$$k^2=\frac{25-\sin^2A}{\cos^2A}$$
So $k$ depends on $A$ and is not a constant.
