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

suppose $\dfrac{SinA}{SinB}=\dfrac{x}{y}=5$
so from right tringle we know $cosA=\sqrt{1-x^{2}}$ and so $$tanA=\dfrac{x}{\sqrt{1-x^{2}}}.$$ the same for y so :
$$\dfrac{tanA}{tanB}=\dfrac{\frac{x}{\sqrt{1-x^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}$$
and we have $x=5y$ so $x^{2}=25y^{2}$ , $1-x^{2}=1-25y^{2}$
$$\dfrac{tanA}{tanB}=\dfrac{\frac{5y}{\sqrt{1-25y^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}=\dfrac{5y\sqrt{1-y^{2}}}{y\sqrt{1-25y^{2}}}$$
now you need to know what is $y$!
