Answer :

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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