Why is $\frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$ valid in a triangle? This is extracted from my module :
In a $\displaystyle\bigtriangleup ABC$,
$$\displaystyle \frac{\sin (A) + \sin(B)}{\sin(A) -\sin(B)} = \frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$$
where $A,B,C$ are the angles and  $a,b,c$ are the opposite sides,I can't figure out how this step holds valid?
 A: The sine rule for a triangle goes as 
$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$, where $R$ is the radius of the circumcircle.
From this, we get $\sin(A) = \frac{a}{2R}$ and $\sin(B) = \frac{b}{2R}$
The problem has $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)}$. Plug in the expressions for $\sin(A)$ and $\sin(B)$ from the above and we get,
$\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{a}{2R}-\frac{b}{2R}} = \frac{\frac{a+b}{2R}}{\frac{a-b}{2R}} = \frac{a+b}{a-b}$
Hence, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$
A: I'm not sure what $k$ is supposed to be, but another way to see that the first and third expressions are equal is to rewrite the first one as
$$\frac{\frac{\sin(A)}{\sin(B)}+1}{\frac{\sin(A)}{\sin(B)}-1}$$
and then note that the law of sines tells you
$$\frac{\sin(A)}{\sin(B)} = \frac{a}{b}$$.
A: 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 $k\neq 0$
$$\dfrac{q+s}{q-s}=\dfrac{k\left( q+s\right) }{k\left( q-s\right) }.$$
A: Geometrically, this is not too hard to see. Presumably, you understand why
$$\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{k(\sin(A) + \sin(B))}{k(\sin(A) - \sin(B))}$$
Let k be the length of the normal from C onto AB. What is then k sin(A) and k sin(B)?
