In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of he equation $b \sin B-c \sin C=a \sin (B-C)$.

$\mathrm{LHS}=k \sin B \sin B-k \sin C \sin C$

$=k\left(\sin ^{2} B-\sin ^{2} C\right)$

$=k[\sin (B+C) \sin (B-C)]$         $\left[\because \sin ^{2} \mathrm{~B}-\sin ^{2} \mathrm{C}=\sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{C})\right]$

$=k[\sin (\pi-\mathrm{A}) \sin (B-C)]$$[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$

$=k \sin A \sin (B-C) \quad[\because a=k \sin A]$

$=a \sin (B-C)=\mathrm{RHS}$

Hence proved.