tan^(−1) (((2x)/(x^2 −1))) + cot^(−1) (((x^2 −1)/(2x))) =((2π)/3) (tan^(−1) (x))^2 +(cot^(−1) (x))^2 =((5π)/8)


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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
	$2)a=tan^(−1) (x) a^2 +{cot^(−1) (tana)}^2 =((5π)/8) a^2 +((π/2)−a)^2 =((5π)/8) a^2 +(π^2 /(4 ))−πa+a^2 =((5π)/8) ((8a^2 −4πa+π^2 )/4)=((5π)/8) 16a^2 −8πa+2π^2 =5π (4a)^2 −2×4a×π+π^2 =5π−π^2 (4a−π)^2 =(5π−π^2 ) a=((π±(√((5π−π^2 )))/4) tan^(−1) x=((π+(√(5π−π^2 )))/4) x=tan(((π+(√(5π−π^2 )))/4))$
	$2) a = {t a n}^{- 1} (x)$ $a^{2} + {{c o t}^{- 1} (t a n a)}^{2} = \frac{5 π}{8}$ $a^{2} + {(\frac{π}{2} - a)}^{2} = \frac{5 π}{8}$ $a^{2} + \frac{π^{2}}{4} - π a + a^{2} = \frac{5 π}{8}$ $\frac{8 a^{2} - 4 π a + π^{2}}{4} = \frac{5 π}{8}$ $16 a^{2} - 8 π a + 2 π^{2} = 5 π$ ${(4 a)}^{2} - 2 \times 4 a \times π + π^{2} = 5 π - π^{2}$ ${(4 a - π)}^{2} = (5 π - π^{2})$ $a = \frac{π \pm \sqrt{(5 π - π^{2}}}{4}$ ${t a n}^{- 1} x = \frac{π + \sqrt{5 π - π^{2}}}{4}$ $x = t a n (\frac{π + \sqrt{5 π - π^{2}}}{4})$
	Answered by TANMAY PANACEA last updated on 14/Oct/20
	$1)x=tana tan^(−1) (((2tana)/(tan^2 a−1)))+tan^(−1) (((2tana)/(tan^2 a−1)))=((2π)/3) 2tan^(−1) {tan(−2a)}=((2π)/3) −4a=((2π)/3)→a=tan^(−1) x=((−π)/6)→x=−(1/( (√3))) tan2θ=((2tanθ)/(1−tan^2 θ))$
	$1) x = t a n a$ ${t a n}^{- 1} (\frac{2 t a n a}{{t a n}^{2} a - 1}) + {t a n}^{- 1} (\frac{2 t a n a}{{t a n}^{2} a - 1}) = \frac{2 π}{3}$ $2 {t a n}^{- 1} {t a n (- 2 a)} = \frac{2 π}{3}$ $- 4 a = \frac{2 π}{3} \to a = {t a n}^{- 1} x = \frac{- π}{6} \to x = - \frac{1}{\sqrt{3}}$ $t a n 2 θ = \frac{2 t a n θ}{1 - {t a n}^{2} θ}$
	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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