cot - 1 ( a b + 1 a - b ) + cot - 1 ( b c + 1 b - c ) + cot - 1 ( c a + 1 c - a ) =

Question

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

Smart Achievers · Accepted Answer

(1)

Ans. 0

Explanation:

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

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

$= \tan^{- 1} a - \tan^{- 1} b + \tan^{- 1} b - \tan^{- 1} c + \tan^{- 1} c - \tan^{- 1} a = 0$

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