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Question

If $\tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = π + \tan^{- 1} (- 7)$ then $x$ is

Moderate

Solution

$I f x > 0, y > 0, x y > 1, t h e n$
$\tan^{- 1} x + \tan^{- 1} y = π + \tan^{- 1} (\frac{x + y}{1 - x y})$
$N o w \tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = π + \tan^{- 1} (\frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{x + 1}{x - 1} \frac{x - 1}{x}}) = π + \tan^{- 1} (\frac{2 x^{2} - x + 1}{1 - x}) = π + \tan^{- 1} (- 7) (given) \Rightarrow \frac{2 x^{2} - x + 1}{1 - x} = - 7 \Rightarrow 2 x^{2} - 8 x + 8 = 0 \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x = 2$

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