Question
Solve for x:
$\tan^{-1}\Bigg(\frac{\text{x - 1}}{\text{x - 2}}\Bigg)+\tan^{-1}\Bigg(\frac{\text{x + 1}}{\text{x + 2}}\Bigg)=\frac{\pi}{4}.$
$\tan^{-1}\Bigg(\frac{\text{x - 1}}{\text{x - 2}}\Bigg)+\tan^{-1}\Bigg(\frac{\text{x + 1}}{\text{x + 2}}\Bigg)=\frac{\pi}{4}.$
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$\cos\theta=\frac{|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2}{2\big|\vec{\text{b}}\big||\vec{\text{c}}|}.$