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

Question

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

Answer 1

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.

Answer 2

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 $y$!