First slide

Properties of ITF

Question

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

Easy

A

1/2
B

-1/2
C

$\frac{1}{\sqrt{2}}$
D

- $\frac{1}{\sqrt{2}}$

Solution

$We have, \tan^{- 1} \frac{x - 1}{x - 2} + \tan^{- 1} \frac{x + 1}{x + 2} = \tan^{- 1} 1 \begin{array}{l} \Rightarrow \tan^{- 1} \frac{x - 1}{x + 1} = \tan^{- 1} 1 - \tan^{- 1} \frac{x + 1}{x + 2} \\ \Rightarrow \tan^{- 1} \frac{x - 1}{x - 2} = \tan^{- 1} \frac{1 - \frac{x + 1}{x + 2}}{1 + \frac{x + 1}{x + 2}} \\ \Rightarrow \tan^{- 1} \frac{x - 1}{x - 2} = \tan^{- 1} \frac{1}{2 x + 3} \end{array} \Rightarrow \frac{x - 1}{x - 2} = \frac{1}{2 x + 3} \Rightarrow (x - 1) (2 x + 3) = x - 2$
⇒ 2x² + x – 3 = x – 2 ⇒ 2x² = 1
$∴ x = \pm \frac{1}{\sqrt{2}}$

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