In a triangle ABC, prove that `(cot(A/2)+cot(B/2)+cot(C/2))/(cotA+cotB+cot(C))=((a+b+c)^2)/(a^2+b^2+c^2)`

Explanation

In `triangleABC`,
`cot(A/2) + cot(B/2) + cot(C/2) = (s(s-a))/(Delta) + (s(s-b))/(Delta) + (s(s-c))/(Delta)= (s(3s-(a+b+c)))/(Delta)`
`=(s(s))/(Delta) = s^(2)/Delta = (2s)^(2)/(4Delta) = (a+b+c)^(2)/(4Delta)`
`cotA + cotB + cotC = (cosA)/(sinA) + (cosB)/(sinB) + (cosC)/(sinC)`
`=(b^(2)+c^(2)-a^(2))/(2bcsinA) +(a^(2)+c^(2)-b^(2))/(2acsinB) + (a^(2)+b^(2)-c^(2))/(2ab sinC)`
`=(b^(2)+c^(2)-a^(2))/(4Delta) + (a^(2)+c^(2)-b^(2))/(4Delta) + (a^(2) + b^(2)-c^(2))/(4Delta)`
`therefore (cot(A/2)+cot(B/2)+cot(C/2))/(cotA + cotB + cotC) = (a+b+c)^(2)/(a^(2)+b^(2)+c^(2))`