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In ` triangle ABC` prove that `(b^2-c^2)/(cos B+cos C) +(c^2-a^2)/(cos C+cos A) +(a^2-b^2)/(cos A + cos B)=0`

Answer» `a/sinA=b/SinB=c/sinC=k`
`a=KsinA,b=KSinB,c=KsinC`
`(b^2-c^2)/(cosB+cosC)=(k^2sin^2B-k^2sin^2c)/(cosB+cosC)=(k^2(1-cos^2B-(1-cos^2C)))/(cosC+cosB)`
`k^2(cos^2C-cos^2B)/(cosB+cosC)=k^2(cosC-CosB)`
LHS=`k^2(cosC-cosB)+k^2(cosA-cosC)+k^2(cosb-cosC)`
`K^2[0]=0=RHS`.`


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