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If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$ then x in is equal to

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Correct option is C

We have,tan−1⁡x−1x+2+tan−1⁡x+1x+2=π4⇒ tan−1⁡x−1x+2+x+1x+21−x−1x+2x+1x+2=π4⇒ 2x(x+2)x2+4+4x−x2+1 =tan⁡π4⇒ 2x(x+2)4x+5 =1⇒ 2x2+4x =4x+5⇒ x =±52

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If tan−1⁡x−1x+2+tan−1⁡x+1x+2=π4 then x in is equal to

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If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$ then x in is equal to