If tan ^{-1} frac{x-1}{x-2}+tan ^{-1} frac{x+ | vedclass

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

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

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(B) Given: $	an ^{-1} \frac{x-1}{x-2}+	an ^{-1} \frac{x+1}{x+2}=\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}$$	an ^{-1} \left[ \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)} ight] = \frac{\pi}{4}$$	an ^{-1} \left[ \frac{x^2+x-2+x^2-x-2}{x^2-4-(x^2-1)} ight] = \frac{\pi}{4}$$	an ^{-1} \left[ \frac{2x^2-4}{-3} ight] = \frac{\pi}{4}$Taking $	an$ on both sides:$\frac{2x^2-4}{-3} = 	an \frac{\pi}{4}$$\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 $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4},$ then find the value of $x$.

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