If any triangle `A B C` , that: `(b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)=0`

Explanation

From sine law, we have,

`a/sinA = b/sinB = c/sinC = k`

`=> a = ksinA, b = ksin B, c = ksinC`

Now,

`L.H.S. = (b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)`

`= (k^2(sin^2B-sin^2C))/(cosB+cosC)+ (k^2(sin^2C-sin^2A))/(cosC+cosA)+ (k^2(sin^2A-sin^2B))/(cosA+cosB)`

`= (k^2((1-cos^2B)-(1-cos^2C)))/(cosB+cosC)+ (k^2((1-cos^2C)-(1-cos^2A)))/(cosC+cosA)+ (k^2((1-cos^2A)-(1-cos^2B)))/(cosA+cosB)`

`=k^2[(cos^2B-cos^2C)/(cosB+cosC)+(cos^2C-cos^2A)/(cosC+cosA)+(cos^2A-cos^2B)/(cosA+cosB)]`

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