For x<−1, if cos−1(x2−11+x2)+sin−1(2x1+x2)−tan−1(2xx2−1)=π3 , Then x is equal to

π−cos−11−x21+x2+sin−12x1+x2+tan−1(2x1−x2)=π3(Given)For x < - 1 π−(−2λ)−(π−2λ)+π+2λ=π3⇒λ=−π3∴x=−3 ]

For x<−1, if cos−1(x2−11+x2)+sin−1(2x1+x2)−tan−1(2xx2−1)=π3 , Then x is equal to

π−cos−11−x21+x2+sin−12x1+x2+tan−1(2x1−x2)=π3(Given)For x < - 1 π−(−2λ)−(π−2λ)+π+2λ=π3⇒λ=−π3∴x=−3 ]

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For $x < - 1,$ if $\cos^{- 1} (\frac{x^{2} - 1}{1 + x^{2}}) + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) - \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{π}{3}$ , Then x is equal to

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$\frac{- 1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

$- \sqrt{3}$

answer is D.

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Detailed Solution

$π - \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + \sin^{- 1} \frac{2 x}{1 + x^{2}} + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3} (G i v e n)$
For $x < - 1$
$π - (- 2 λ) - (π - 2 λ) + π + 2 λ = \frac{π}{3} \Rightarrow λ = - \frac{π}{3}$
$∴ x = - \sqrt{3}$ ]