tan-1-2x-x-2-1-cot-1-x-2-1-2x-2pi-3-tan-1-x-2-cot-1-x-2-5pi-8-

Question Number 117926 by bemath last updated on 14/Oct/20

\tan^{- 1} (\frac{2 x}{x^{2} - 1}) + \cot^{- 1} (\frac{x^{2} - 1}{2 x}) = \frac{2 π}{3}

{(\tan^{- 1} (x))}^{2} + {(\cot^{- 1} (x))}^{2} = \frac{5 π}{8}

Answered by TANMAY PANACEA last updated on 14/Oct/20

2) a = {t a n}^{- 1} (x)

a^{2} + {{c o t}^{- 1} (t a n a)}^{2} = \frac{5 π}{8}

a^{2} + {(\frac{π}{2} - a)}^{2} = \frac{5 π}{8}

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

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

16 a^{2} - 8 π a + 2 π^{2} = 5 π

{(4 a)}^{2} - 2 \times 4 a \times π + π^{2} = 5 π - π^{2}

{(4 a - π)}^{2} = (5 π - π^{2})

a = \frac{π \pm \sqrt{(5 π - π^{2}}}{4}

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

x = t a n (\frac{π + \sqrt{5 π - π^{2}}}{4})

Answered by TANMAY PANACEA last updated on 14/Oct/20

1) x = t a n a

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

2 {t a n}^{- 1} {t a n (- 2 a)} = \frac{2 π}{3}

- 4 a = \frac{2 π}{3} \to a = {t a n}^{- 1} x = \frac{- π}{6} \to x = - \frac{1}{\sqrt{3}}

t a n 2 θ = \frac{2 t a n θ}{1 - {t a n}^{2} θ}

Commented by bemath last updated on 14/Oct/20

thank you sir

Answered by bobhans last updated on 14/Oct/20

(1) case (1) for \frac{2 x}{x^{2} - 1} > 0 \Rightarrow \tan^{- 1} (\frac{2 x}{x^{2} - 1}) + \cot^{- 1} (\frac{x^{2} - 1}{2 x}) = \frac{2 π}{3}

\Rightarrow {2 \tan}^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3}; \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{π}{3}

\Rightarrow \frac{2 x}{x^{2} - 1} = \sqrt{3}; \sqrt{3} x^{2} - 2 x - \sqrt{3} = 0

\Rightarrow x = \frac{2 \pm \sqrt{4 + 12}}{2 \sqrt{3}} = \frac{2 \pm 4}{2 \sqrt{3}}

\Rightarrow x = \sqrt{3} or x = - \frac{1}{\sqrt{3}}

case (2) for \frac{2 x}{x^{2} - 1} < 0 \Rightarrow \tan^{- 1} (\frac{2 x}{x^{2} - 1}) + \cot^{- 1} (\frac{x^{2} - 1}{2 x}) = \frac{2 π}{3}

\Rightarrow \tan^{- 1} (\frac{2 x}{x^{2} - 1}) + π + \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3}

\Rightarrow {2 \tan}^{- 1} (\frac{2 x}{x^{2} - 1}) = - \frac{π}{3}; \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = - \frac{π}{6}

\Rightarrow \frac{2 x}{x^{2} - 1} = - \frac{1}{\sqrt{3}}; x^{2} + 2 \sqrt{3} x - 1 = 0

\Rightarrow x = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2} = \frac{- 2 \sqrt{3} \pm 4}{2}

\Rightarrow x = - \sqrt{3} + 2 or x = - \sqrt{3} - 2

Commented by bemath last updated on 14/Oct/20

gave kudos

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