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Question

Prove b2c2a2sin2A+c2a2b2sin2B+a2b2c2sin2C=0

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Solution

We know that in a triangle a=ksinA,b=ksinB,c=ksinC and cosA=b2+c2a22bc and similarly using it for cosB andcosC

So,sin2Aa2=2a2sinAcosA=2aka2b2+c2a22bc=b2+c2a22kabc

sin2Bb2=2b2sinBcosB=2bkb2c2+a2b22ca=c2+a2b22kabc

sin2Cc2=2c2sinCcosC=2ckc2a2+b2c22ab=a2+b2c22kabc

Thus,
L.H.S=(b2c2)sin2Aa2+(c2a2)sin2Bb2+(a2b2)sin2Cc2

=(b2c2)b2+c2a22kabc+(c2a2)c2+a2b22kabc+(a2b2)a2+b2c22kabc

=12akbc[(b2c2)(b2+c2a2)+(c2a2)(c2+a2b2)+(a2b2)(a2+b2c2)]

=12akbc[b4c4a2(b2c2)+c4a4b2(c2a2)+a4b4c2(a2b2)]

=12akbc[a2b2+a2c2b2c2+b2a2c2a2+c2b2]

=12akbc(0)=0=R.H.S
Hence proved

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