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Question

Prove that a2sin(BC)sinB+sinC+b2sin(CA)sinC+sinA+c2sin(AB)sinA+sinB=0

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Solution

Given:
We know that,
a=2RsinA and A+B+C=π

A=π(B+C)

sinA=sin(π(B+C))=sin(B+C)

a2sin(BC)sinB+sinC=4R2sin2Asin(BC)sinB+sinC (using the formula in triangle)

=4R2sinAsin(B+C)sin(BC)sinB+sinC

=4R2sinA(sin2Bsin2C)sinB+sinC

=4R2sinA(sinBsinC)

Similarly, b2sin(CA)sinC+sinA=4R2sinB(sinCsinA)

and

c2sin(AB)sinA+sinB=4R2sinC(sinAsinB)

Adding we get a2sin(BC)sinB+sinC+b2sin(CA)sinC+sinA+c2sin(AB)sinA+sinB=0

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