Need to prove: a² sin ( B – C ) = (b² – c²) sin A

We know that, where R is the circumradius.

Therefore,

a = 2R sinA ---- (a)

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

From Right hand side,

(b² – c²) sin A

{(2R sinB)² – (2R sinC)²} sinA

4R²( sin²B – sin²C )sinA

We know, sin²B – sin²C = sin(B + C)sin(B – C)

So,

4R²(sin(B + C)sin(B – C))sinA

4R²(sin( – A)sin(B – C))sinA [ As, A + B + C = ]

4R²(sinAsin(B – C))sinA [ As, sin( – ) = sin ]

4R²sin²A sin(B – C)

a²sin(B – C) [From (a)]

Left hand side. [Proved]