The value of ${tan}^{- 1} 1 + {tan}^{- 1} 2 + {tan}^{- 1} 3$ is-

0

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1

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π

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- π

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Solution

The correct option is C $π$
The value of ${tan}^{- 1} 1 + {tan}^{- 1} 2 + {tan}^{- 1} 3$
$= π + {tan}^{- 1} (\frac{1 + 2}{1 - 2}) + {tan}^{- 1} 3 = π + {tan}^{- 1} (- 3) + {tan}^{- 1} 3 = π - {tan}^{- 1} 3 + {tan}^{- 1} 3 = π$

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Similar questions

{tan}^{- 1} (1) + {tan}^{- 1} (2) + {tan}^{- 1} (3) = π

{tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A) = {\begin{matrix} 0, i f \frac{π}{4} < A < \frac{π}{2} π, i f 0 < A < \frac{π}{2} \end{matrix}

({tan}^{- 1} π + {tan}^{- 1} (\frac{1}{π})) + {tan}^{- 1} \sqrt{3} - {sec}^{- 1} (- 2)

Find the value of $t a n^{- 1} (- \frac{1}{\sqrt{3}}) + c o t^{- 1} (\frac{1}{\sqrt{3}}) + t a n^{- 1} [s i n (\frac{- π}{2})] .$

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} (\frac{1}{2} t a n 2 A) + {t a n}^{- 1} (c o t A) + {t a n}^{- 1} ({c o t}^{3} A)

0

π / 4 < A < π / 2

= π

0 < A < π / 4

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