If tan ^{-1} 2x + tan ^{-1} 3x = frac{pi}{4}, | vedclass

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(C) Given the equation: $	an ^{-1} 2x + 	an ^{-1} 3x = \frac{\pi}{4}$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{

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$\frac{1}{6}$

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(C) Given the equation: $	an ^{-1} 2x + 	an ^{-1} 3x = \frac{\pi}{4}$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$,we get: $	an ^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} ight) = \frac{\pi}{4}$ $	an ^{-1} \left( \frac{5x}{1 - 6x^2} ight) = \frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{5x}{1 - 6x^2} = 	an \left( \frac{\pi}{4} ight) = 1$ $5x = 1 - 6x^2$ $6x^2 + 5x - 1 = 0$ Factoring the quadratic equation: $6x^2 + 6x - x - 1 = 0$ $6x(x + 1) - 1(x + 1) = 0$ $(6x - 1)(x + 1) = 0$ This gives $x = \frac{1}{6}$ or $x = -1$. Since $	an ^{-1} 2x + 	an ^{-1} 3x = \frac{\pi}{4} > 0$,$x$ must be positive. Therefore,$x = -1$ is rejected. Thus,$x = \frac{1}{6}$.

If $\tan ^{-1} 2x + \tan ^{-1} 3x = \frac{\pi}{4}$,then $x = $

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