If a>b>c>0,then cot−1ab+1a−b+cot−1bc+1b−c+cot−1ca−1c−a
0
π2
π
3π2
Given a>b>c>0. Now cot-1ab+1a-b+cot-1bc+1b-c+cot-1ca+1c-a=cot-1ab+1a-b+cot-1bc+1b-c+cot-1-ac+1a-c =cot-1a-cot-1b+cot-1b-cot-1c+π-cot-1a-cot-1c ∵if x>y, then cot-1xy+1x-y=cot-1x-cot-1y, also cot-1-x=π-cot-1x =π