If a>b>c>0thencot−1ab+1a−b+cot−1bc+1b−c+cot−1ca+1c−a=

Since a>b>c>0, we have a-b>0, b-c>0 and c-a<0  ∴cot−1ab+1a−b+cot−1bc+1b−c+cot−1ca+1c−a=cot−1ab+1a−b+cot−1bc+1b−c+cot−1-ca+1a-c                                                                                          =cot-1a-cot-1b+cot-1b-cot-1c+π-cot-1a-cot-1c                                                                                                                                       ∵cot-1-x=π-cot-1x                                                                                          =π