usig the identity::
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$
we conclude:
$\tan ^{-1} \frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}=\frac{\pi}{4}$
$\tan ^{-1} \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=\frac{\pi}{4}$
$\tan ^{-1} \frac{x^{2}+2 x-x-2+x^{2}-2 x+x-2}{x^{2}-4-\left(x^{2}-1\right)}=\frac{\pi}{4}$
$\frac{2 x^{2}-4}{x^{2}-4-x^{2}+1}=\tan \left(\frac{\pi}{4}\right)$
$\left(2 \times^{\wedge}-4\right) /-3=1$
$2 x^{-2}=1$
therefore, the value of $x$ is either $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}$