The value of ${tan}^{- 1} 2 + {tan}^{- 1} 3 + {tan}^{- 1} 4$ is equal to

π - {tan}^{- 1} \frac{3}{5}

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π + {tan}^{- 1} \frac{5}{3}

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π - {tan}^{- 1} \frac{5}{3}

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π + {tan}^{- 1} \frac{3}{5}

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Solution

The correct option is D $π + {tan}^{- 1} \frac{3}{5}$
${tan}^{- 1} 2 + {tan}^{- 1} 3 + {tan}^{- 1} 4 = π + {tan}^{- 1} (\frac{2 + 3}{1 - 2.3}) + {tan}^{- 1} 4 (∵ 2.3 > 1) = π - {tan}^{- 1} 1 + {tan}^{- 1} 4 = π + {tan}^{- 1} (\frac{4 - 1}{1 + 1.4}) = π + {tan}^{- 1} \frac{3}{5}$

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Similar questions

{tan}^{- 1} \frac{3}{4} + {tan}^{- 1} \frac{3}{5} - {tan}^{- 1} \frac{8}{19} = \frac{π}{4}

{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}

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{t a n}^{- 1} \frac{1}{2} + {t a n}^{- 1} \frac{1}{3} = \frac{π}{4}

{t a n}^{- 1} \frac{1}{2} + {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

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

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{sin}^{- 1} \frac{4}{5} + {sin}^{- 1} \frac{3}{5} = \frac{π}{2}

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

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