Q.In $\triangle ABC$ prove that $a\sin A - b\sin B = c\sin(A-B)$.

'Solving a triangle' means finding every one of its six elements (three sides, three angles) once any three are known, provided at least one of the three known elements is a side (angles alone only fix the shape, not the size). The usual notation labels each side by the lower-case letter of the vertex it is opposite: $a=BC$, $b=CA$, $c=AB$. The Sine Rule is the first master relation for this: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$, where $R$ is the circumradius of the triangle. It is proved in two stages — first that the three ratios $a/\sin A$, $b/\sin B$, $c/\sin C$ are equal to each other, using the triangle's area computed via each pair of sides and their included angle; and second that this common rati …