<s>[INST] 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?
 [/INST] 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, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$
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