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

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\frac{π}{2}

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π

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None

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Solution

The correct option is C $π$
$t a n^{- 1} + t a n^{- 1} 2 = π + t a n^{- 1} \frac{1 + 2}{1 - 1 \times 2} = π - t a n^{- 1} 3 \Rightarrow t a n^{- 1} 1 + t a n^{- 1} 2 + t a n^{- 1} 3 = π$

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Similar questions

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

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})

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

{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}{21} + {t a n}^{- 1} \frac{1}{13} + {t a n}^{- 1} (- \frac{1}{8}) = 0

{t a n}^{- 1} \frac{2}{3} = \frac{1}{2} {t a n}^{- 1} \frac{12}{5}

{t a n}^{- 1} \frac{1}{4} + {t a n}^{- 1} \frac{2}{9} = \frac{1}{2} {c o s}^{- 1} \frac{3}{5}

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