In $\triangle ABC$ prove that $\;$ $\dfrac{\sin B}{\sin C} = \dfrac{c - a \cos B}{b - a \cos C}$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\begin{aligned}
RHS & = \dfrac{c - a \cos B}{b - a \cos C} \\\\
& = \dfrac{c - a \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right)}{b - a \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)} \\\\
& = \dfrac{\dfrac{c^2 + b^2 - a^2}{c}}{\dfrac{b^2 + c^2 - a^2}{b}} \\\\
& = \dfrac{b}{c} \\\\
& = \dfrac{\sin B}{\sin C} = LHS
\end{aligned}$