Solve: $\;$ $\tan^{-1} \left(\dfrac{x - 1}{x - 2}\right) + \tan^{-1} \left(\dfrac{x + 1}{x + 2}\right) = \dfrac{\pi}{4}$
$\tan^{-1} \left(\dfrac{x - 1}{x - 2}\right) + \tan^{-1} \left(\dfrac{x + 1}{x + 2}\right) = \dfrac{\pi}{4}$
i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{x - 1}{x - 2} + \dfrac{x + 1}{x + 2}}{1 - \left(\dfrac{x - 1}{x - 2}\right) \times \left(\dfrac{x + 1}{x + 2}\right)}\right] = \dfrac{\pi}{4}$ $\;\;\;$ provided $\left(\dfrac{x - 1}{x - 2}\right) \times \left(\dfrac{x + 1}{x + 2}\right) < 1$
i.e. $\;$ $\tan^{-1} \dfrac{\left(x - 1\right) \left(x + 2\right) + \left(x - 2\right) \left(x + 1\right)}{x^2 - 4 - x^2 + 1} = \dfrac{\pi}{4}$
i.e. $\;$ $\dfrac{x^2 + x - 2 + x^2 - x - 2}{- 3} = \tan \left(\dfrac{\pi}{4}\right)$
i.e. $\;$ $\dfrac{2 x^2 - 4}{- 3} = 1$
i.e. $\;$ $2 x^2 = 1$ $\implies$ $x^2 = \dfrac{1}{2}$ $\implies$ $x = \pm \dfrac{1}{\sqrt{2}}$
Verification:
Putting $x = + \dfrac{1}{\sqrt{2}}$ in the given equation,
$LHS = \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}} - 2} \right) + \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}} + 1}{\dfrac{1}{\sqrt{2}} + 2} \right)$
$= \tan^{-1} \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) + \tan^{-1} \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right)$
$= \tan^{-1} \left[\dfrac{\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}} + \dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}}{1 - \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right)}\right]$ $\;\;\;$ $\because \; \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) < 1$
$= \tan^{-1} \left(\dfrac{1 + 2 \sqrt{2} - \sqrt{2} - 4 + 1 - 2 \sqrt{2} + \sqrt{2} - 4}{1 - 8 - 1 + 2}\right)$
$= \tan^{-1} \left(\dfrac{- 6}{- 6}\right) = \tan^{-1} \left(1\right) = \dfrac{\pi}{4} = RHS$
Putting $x = - \dfrac{1}{\sqrt{2}}$ in the given equation,
$LHS = \tan^{-1} \left(\dfrac{- \dfrac{1}{\sqrt{2}} - 1}{- \dfrac{1}{\sqrt{2}} - 2} \right) + \tan^{-1} \left(\dfrac{- \dfrac{1}{\sqrt{2}} + 1}{- \dfrac{1}{\sqrt{2}} + 2} \right)$
$= \tan^{-1} \left(\dfrac{- 1 - \sqrt{2}}{- 1 - 2 \sqrt{2}}\right) + \tan^{-1} \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right)$
$= \tan^{-1} \left[\dfrac{\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}} + \dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}}{1 - \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right)}\right]$ $\;\;\;$ $\because \; \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right) < 1$
$= \tan^{-1} \left(\dfrac{- 1 + 2 \sqrt{2} - \sqrt{2} + 4 - 1 - 2 \sqrt{2} + \sqrt{2} + 4}{8 - 1 - 2 + 1}\right)$
$= \tan^{-1} \left(\dfrac{6}{6}\right) = \tan^{-1} \left(1\right) = \dfrac{\pi}{4} = RHS$
$\therefore \;$ The solution set is $\left\{\pm \dfrac{1}{\sqrt{2}} \right\}$