Question
Solve for x:$\tan^{-1}\Bigg(\frac{\text{x - 1}}{\text{x - 2}}\Bigg)+\tan^{-1}\Bigg(\frac{\text{x + 1}}{\text{x + 2}}\Bigg)=\frac{\pi}{4}.$
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$\text{x}_\text{i}$
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$-5$
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$-4$
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$1$
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$2$
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$\text{p}_\text{i}$
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$\frac{1}{4}$
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$\frac{1}{8}$
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$\frac{1}{2}$
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$\frac{1}{8}$
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