If $$0 < a < b < c,$$    then $${\cot ^{ - 1}}\left( {\frac{{ab + 1}}{{a - b}}} \right) + {\cot ^{ - 1}}\left( {\frac{{bc + 1}}{{b - c}}} \right) + {\cot ^{ - 1}}\left( {\frac{{ca + 1}}{{c - a}}} \right) = $$

Solution :
$$\eqalign{ & \because a - b < 0,{\text{ so}} \cr & {\cot ^{ - 1}}\frac{{ab + 1}}{{a - b}} = {\cot ^{ - 1}}b - {\cot ^{ - 1}}a + \pi \cr & b - c < 0,{\text{ so }}{\cot ^{ - 1}}\frac{{bc + 1}}{{b - c}} = {\cot ^{ - 1}}c - {\cot ^{ - 1}}b + \pi \cr & c - a > 0,{\text{ so }}{\cot ^{ - 1}}\frac{{ca + 1}}{{c - a}} = {\cot ^{ - 1}}a - {\cot ^{ - 1}}c \cr & {\text{Adding we get,}} \cr & {\cot ^{ - 1}}\frac{{ab + 1}}{{a - b}} + {\cot ^{ - 1}}\frac{{bc + 1}}{{b - c}} + {\cot ^{ - 1}}\frac{{ca + 1}}{{c - a}} = 2\pi \cr} $$