If tan ^{-1}left(frac{x-1}{x-2}right)+tan ^{- | vedclass

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(C) Given the equation: $	an ^{-1}\left(\frac{x-1}{x-2}ight)+	an ^{-1}\left(\frac{x+1}{x+2}ight)=\frac{\pi}{4}$ Using the formula $	an ^{-1} A

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$\pm \frac{1}{\sqrt{2}}$

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(C) Given the equation: $	an ^{-1}\left(\frac{x-1}{x-2}ight)+	an ^{-1}\left(\frac{x+1}{x+2}ight)=\frac{\pi}{4}$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$: $	an ^{-1} \left[ \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left( \frac{x-1}{x-2} ight) \left( \frac{x+1}{x+2} ight)} ight] = \frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = 	an \frac{\pi}{4} = 1$ Expanding the terms: $\frac{(x^2+x-2) + (x^2-x-2)}{(x^2-4) - (x^2-1)} = 1$ $\frac{2x^2 - 4}{-4 + 1} = 1$ $\frac{2x^2 - 4}{-3} = 1$ $2x^2 - 4 = -3$ $2x^2 = 1$ $x^2 = \frac{1}{2}$ $x = \pm \frac{1}{\sqrt{2}}$

If $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$,then the values of $x$ are

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