Inverse Trigonometric Functions — Class 12 Maths Solution

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

$\Rightarrow$ ${\tan ^{ - 1}}\left[ {\frac{{\frac{{x - 1}}{{x - 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x - 2}}} \right)\left( {\frac{{x + 1}}{{x + 2}}} \right)}}} \right] = \frac{\pi }{4}$

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

$\Rightarrow$ $\frac{{{x^2} + 2x - x - 2 + {x^2} - 2x + x - 2}}{{{x^2} - 4 - {x^2} + 1}} = \tan \frac{\pi }{4}$

$\Rightarrow$ $\frac{{2{x^2} - 4}}{{ - 3}} = 1 \Rightarrow 2{x^2} = - 3 + 4 \Rightarrow {x^2} = \frac{1}{2} \Rightarrow x = \pm \frac{1}{{\sqrt 2 }}$