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 in is equal to

Moderate

A

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

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

$\pm \sqrt{\frac{5}{2}}$
D

$\pm \frac{1}{2}$

Solution

We have,
$\tan^{- 1} \frac{x - 1}{x + 2} + \tan^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}$
$\Rightarrow \tan^{- 1} [\frac{\frac{x - 1}{x + 2} + \frac{x + 1}{x + 2}}{1 - (\frac{x - 1}{x + 2}) (\frac{x + 1}{x + 2})}] = \frac{π}{4}$
$\begin{array}{lrl} \Rightarrow [\frac{2 x (x + 2)}{x^{2} + 4 + 4 x - x^{2} + 1}] = \tan \frac{π}{4} \\ \Rightarrow \frac{2 x (x + 2)}{4 x + 5} = 1 \\ \Rightarrow 2 x^{2} + 4 x = 4 x + 5 \\ \Rightarrow x = \pm \sqrt{\frac{5}{2}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App