# 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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