Solve for x: tan⁻¹((x - 1)/(x - 2)) + tan⁻¹((x + 1)/(x + 2)) = π/4

Using $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$: $\frac{a+b}{1-ab} = \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)} = \frac{2x^2-4}{-3}$ So $\tan^{-1}\left(\frac{2x^2-4}{-3}\right) = \frac{\pi}{4}$, giving $\frac{2x^2-4}{-3} = \tan\frac{\pi}{4} = 1$. $2x^2 - 4 = -3$, so $x^2 = \frac{1}{2}$. For $x = \pm\frac{1}{\sqrt2}$, $ab = \frac{x^2-1}{x^2-4} = \frac{1}{7} < 1$, so the formula applies and both values satisfy the equation. Hence $x = \pm\frac{1}{\sqrt{2}}$.