In triangle ABC, prove the following:

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

Then

Consider the RHS of the equation $a^{2} \sin (B-C)=\left(b^{2}-c^{2}\right) \sin A$.

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

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

$=k^{2} \sin A[\sin (\pi-A) \sin (B-C)] \quad[\because A+B+C=\pi]$

$=k^{2} \sin A[\sin (A) \sin (B-C)]$

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

$=a^{2} \sin (B-C)=$ LHS $\quad[\because a=k \sin A]$

hence proved.