From the $\sin $ rule, you know that
$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$
In a proportion such as
$$\frac{p}{q}=\frac{r}{s},$$
we have (by a property of proportions, which you can check algebraically):
$$\frac{p+r}{p-r}=\frac{q+s}{q-s}.$$
Setting $p=a$, $q=\sin A$, $r=b$, $s=\sin B$ gives you
$$\frac{a+b}{a-b}=\frac{\sin A+\sin B}{\sin A-\sin B}.$$
Added: Algebraic deduction:
$$\dfrac{p+r}{p-r}=\dfrac{q+s}{q-s}\Leftrightarrow \left( p+r\right) \left( q-s\right) -\left( p-r\right)\left( q+s\right) =0$$
$$\Leftrightarrow 2qr-2ps=0\Leftrightarrow qr=ps\Leftrightarrow \dfrac{p}{q}=\dfrac{r}{s}.$$
And clearly, for $k\neq 0$
$$\dfrac{q+s}{q-s}=\dfrac{k\left( q+s\right) }{k\left( q-s\right) }.$$