# How do you prove \frac{a-c}{c} = \frac{\sin A - \sin C}{\sin C} using the law of sines?

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C = R$
so, $a = R \sin A , b = R \sin B$ & $c = R \sin C$
So, $\frac{a - c}{c} = \frac{R \sin A - R \sin C}{R \sin C} = \frac{R \left(\sin A - \sin C\right)}{R \sin C} = \frac{\sin A - \sin C}{\sin} C$