# In any ΔABC, prov

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]

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