Find the value of ${tan}^{- 1} 1 + {tan}^{- 1} 2 + {tan}^{- 1} 3$ .

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Solution

Now,
${tan}^{- 1} 1 + {tan}^{- 1} 2 + {tan}^{- 1} 3$
$= {tan}^{- 1} 1 + {tan}^{- 1} \frac{2 + 3}{1 - 6}$
$= \frac{π}{4} + {tan}^{- 1} (- 1)$
$= \frac{π}{4} + \frac{3 π}{4}$
$= π$ .

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Similar questions

{tan}^{- 1} 1 + {tan}^{- 1} 2 + {tan}^{- 1} 3

- 1 \frac{1}{7} + t a n^{-} 1 \frac{1}{13} = t a n^{-} 1 \frac{2}{9}

{cot}^{- 1} 1 - {tan}^{- 1} 2 + {cot}^{- 1} 3 .

t a n^{- 1} 1 + t a n^{- 1} 2 + t a n^{- 1} 3 = π

{tan}^{- 1} (\frac{1}{7}) + {tan}^{- 1} (\frac{1}{13}) = {tan}^{- 1} (\frac{2}{9})

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