Question

In Triangle ABC prove that cotA/2+cotB/2+cotC/2=a+b+c/b+c-a*cotA/2

Answer 1

A + B + C = π
=> A/2 + B/2 = π/2 - C/2
=> tan (A/2 + B/2) = tan (π/2 - C/2)
=> (tan A/2 + tanB/2) / (1 - tan A/2 tan B/2) = cot C/2
=> (cot A/2 + cot B/2) / (cot A/2 cot B/2 - 1) = cot C/2
=> cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2
Of course its only true when ABC are angles of a triangle