Trigonometric Functions — Maths STD 12 Science — Question

Given, $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\cot ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$;
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)=\frac{\pi}{4}-\tan ^{-1}\left(\frac{x+1}{x+2}\right)$
$=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{x+1}{x+2}\right)$
$=\tan ^{-1}\left[\frac{1-\frac{x+1}{x+2}}{1+\frac{x+1}{x+2}}\right] \ldots \ldots \ldots .\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$=\tan ^{-1}\left[\frac{x+2-x-1}{x+2+x+1}\right]$
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)=\tan ^{-1}\left(\frac{1}{2 x+3}\right)$
$\frac{x-1}{x-2}=\frac{1}{2 x+3}$
$(x-1)(2 x+3)=x-2$
$2 x^2-1=0$
$2 x^2=1$
$x^2=\frac{1}{2}$
$x= \pm \frac{1}{\sqrt{2}}$