If tan - 1 ( x - 1 x + 2 ) + tan - 1 ( x + 1 x + 2 ) = 4 , then x =

Question

If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4},$ then $x =$

Smart Achievers · Accepted Answer

(3)

Ans. $\pm \sqrt{\frac{5}{2}}$

Explanation:

$\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$

$\Rightarrow$ $\tan^{- 1} (\frac{\frac{x - 1}{x + 2} + \frac{x + 1}{x + 2}}{1 - \frac{(x - 1) (x + 1)}{{(x + 2)}^{2}}}) = \frac{π}{4}$

$\Rightarrow$ $\frac{(x - 1 + x + 1) (x + 2)}{{(x + 2)}^{2} - (x^{2} - 1)} = 1$

$\Rightarrow$ $\frac{(2 x) (x + 2)}{x^{2} + 4 x + 4 - x^{2} + 1} = 1$

$\Rightarrow$ $2 x^{2} + 4 x = 4 x + 5$

$\Rightarrow$ $2 x^{2} = 5$

$\Rightarrow$ $x^{2} = \frac{5}{2}$

$∴$ $x = \pm \sqrt{\frac{5}{2}}$

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