We have,
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4} \Rightarrow \tan$ $^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4} \\$
${\left[\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}\right]=\tan$ \frac{\pi}{4} \Rightarrow \frac{x^2+x-2+x^2-x-2}{x^2-4-x^2+1}=1 \quad\left[\because \tan \frac{\pi}{4}=1\right]} \\$
$\frac{2 x^2-4}{-3}=1 \Rightarrow 2 x^2-4=-3 \Rightarrow 2 x^2=-3+4=1 \Rightarrow 2 x^2=1 \Rightarrow x^2=\frac{1}{2} \Rightarrow x= \pm \frac{1}{\sqrt{2}} \\$
