$t a n^{- 1} 1 + t a n^{- 1} 2 + t a n^{- 1} 3$ is equal to

\frac{π}{2}

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π

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2 π

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\frac{26 π}{24}

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Solution

The correct option is B $π$
Given ${tan}^{- 1} (1) + {tan}^{- 1} (2) + {tan}^{- 1} (3)$

$= π + {tan}^{- 1} (\frac{3}{1 - 2}) + {tan}^{- 1} (3)$ since ${tan}^{- 1} (x) + {tan}^{- 1} (y) = π + {tan}^{- 1} (\frac{x + y}{1 - x y})$ as $x y > 1$

$= π + {tan}^{- 1} (- 3) + {tan}^{- 1} (3)$

$= π - {tan}^{- 1} (3) + {tan}^{- 1} (3)$

$= π$

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