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If $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}$, then find the value of $x$.
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L.H.S. $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)$
$=\tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]$
$=\tan ^{-1}\left[\frac{2 x^2-4}{-3}\right]$
R.H.S. $=\frac{\pi}{4}=\tan ^{-1}(1)$
$\therefore \quad \frac{2 x^2-4}{-3}=1 \quad$ or $\quad 2 x^2-4=-3$
or $2 x^2=4-3=1 \Rightarrow x^2=\frac{1}{2} \Rightarrow x=\pm \frac{1}{\sqrt{2}}$
$=\tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]$
$=\tan ^{-1}\left[\frac{2 x^2-4}{-3}\right]$
R.H.S. $=\frac{\pi}{4}=\tan ^{-1}(1)$
$\therefore \quad \frac{2 x^2-4}{-3}=1 \quad$ or $\quad 2 x^2-4=-3$
or $2 x^2=4-3=1 \Rightarrow x^2=\frac{1}{2} \Rightarrow x=\pm \frac{1}{\sqrt{2}}$
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