In any ΔABC, prove that

Solution:

Need to prove: $a^{2} \sin (B-C)=\left(b^{2}-c^{2}\right) \sin A$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where R is the circumradius.

Therefore,

$a=2 R \sin A \ldots(a)$

Similarly, b = 2R sinB and c = 2R sinC

From Right hand side,

$=\left(b^{2}-c^{2}\right) \sin A$

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

$=4 R^{2}\left(\sin ^{2} B-\sin ^{2} C\right) \sin A$

We know, $\sin ^{2} B-\sin ^{2} C=\sin (B+C) \sin (B-C)$

So,

$=4 \mathrm{R}^{2}(\sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}$

$=4 \mathrm{R}^{2}(\sin (\pi-\mathrm{A}) \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}[\mathrm{As}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$

$=4 \mathrm{R}^{2}(\sin \mathrm{A} \sin (\mathrm{B}-\mathrm{C})) \sin \mathrm{A}[\mathrm{As}, \sin (\pi-\theta)=\sin \theta]$

$=4 \mathrm{R}^{2} \sin ^{2} \mathrm{~A} \sin (\mathrm{B}-\mathrm{C})$

$=\mathrm{a}^{2} \sin (\mathrm{B}-\mathrm{C})[$ From $(\mathrm{a})]$

$=$ Left hand side. [Proved]