If a, b, c are distinct non-zero real numbers having the same sign, then prove that
${cot}^{- 1} (\frac{a b + 1}{a - b}) + {cot}^{- 1} (\frac{b c + 1}{b - c}) + {cot}^{- 1} (\frac{c a + 1}{c - a}) = π o r 2 π$

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Solution

${cot}^{- 1} (\frac{a b + 1}{a - b}) + {cot}^{- 1} (\frac{b c + 1}{b - c}) + {cot}^{- 1} (\frac{a c + 1}{c - a}) = π$

Take LHS
$= {cot}^{- 1} (\frac{a b + 1}{a - b}) + {cot}^{- 1} (\frac{b c + 1}{b - c}) + {cot}^{- 1} (\frac{a c + 1}{c - a})$

$= {tan}^{- 1} (\frac{a - b}{1 + a b}) + {tan}^{- 1} (\frac{b - c}{1 + b c}) + {tan}^{- 1} (\frac{c - a}{1 + a c})$

$= π + {tan}^{- 1} (a) - {tan}^{- 1} (b) + {tan}^{- 1} (b) + {tan}^{- 1} (c) + {tan}^{- 1} (c) - {tan}^{- 1} (a)$

$= π$

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