1.

Prove that `(a^2sin(B-C))/(sinb+sinC)+(b^2"sin"(C-A))/(sinC+sinA)+(c^2"sin"(A-B))/(sinA+sinB)=0`

Answer» `(a^2sin(B-C))/(sinB+sinC) = ((2RsinA)^2 sin(B-C))/(sinB+sinC)`
`=(4R^2sin^2Asin(B-C))/(sinB+sinC)`
`=(4R^2sinA*sinAsin(B-C))/(sinB+sinC)`
`=(4R^2sinA*sin(pi-(B+C)sin(B-C))/(sinB+sinC)`
`=(4R^2sinA*sin(B+C)sin(B-C))/(sinB+sinC)`
`=(4R^2sinA*(sin^2B-sin^2C))/(sinB+sinC)`
`=(4R^2sinA*(sinB+sinC)(sinB-sinC))/(sinB+sinC)`
`=4R^2sinA(sinB-sinC)`
Similarly,
`(b^2sin(C-A))/(sinC+sinA) =4R^2sinB(sinC-sinA)`
`(c^2sin(A-B))/(sinA+sinB) =4R^2sinC(sinA-sinB)`
`:. L.H.S. = (a^2sin(B-C))/(sinB+sinC) +(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB) = 4R^2[sinAsinB -sinAsinC+sinBsinC-sinBsinA+sinCsinA-sinCsinB]`
`=4R^2[0] = 0 = R.H.S.`


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