If a>b>c>0,then cot−1ab+1a−b+cot−1bc+1b−c+cot−1ca−1c−a

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

If a>b>c>0,then cot−1ab+1a−b+cot−1bc+1b−c+cot−1ca−1c−a

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

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If $a > b > c > 0,$ then $\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a - 1}{c - a})$

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$π$

$\frac{3 π}{2}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$G i v e n a > b > c > 0 . N o w c o t^{- 1} (\frac{a b + 1}{a - b}) + c o t^{- 1} (\frac{b c + 1}{b - c}) + c o t^{- 1} (\frac{c a + 1}{c - a}) = c o t^{- 1} (\frac{a b + 1}{a - b}) + c o t^{- 1} (\frac{b c + 1}{b - c}) + c o t^{- 1} (- (\frac{a c + 1}{a - c})) = c o t^{- 1} a - c o t^{- 1} b + c o t^{- 1} b - c o t^{- 1} c + π - (\cot^{- 1} a - \cot^{- 1} c) (∵ if x > y, then \cot^{- 1} (\frac{xy + 1}{x - y}) = \cot^{- 1} x - \cot^{- 1} y, also \cot^{- 1} (- x) = π - \cot^{- 1} x) = π$