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

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(B) Given 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} ig

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$\frac{1}{6}$

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(B) Given 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}$$	an^{-1} \left( \frac{5x}{1 - 6x^2} ight) = 	an^{-1}(1)$$\frac{5x}{1 - 6x^2} = 1$$1 - 6x^2 = 5x$$6x^2 + 5x - 1 = 0$$(6x - 1)(x + 1) = 0$So,$x = \frac{1}{6}$ or $x = -1$.If $x = -1$,then $	an^{-1}(-2) + 	an^{-1}(-3) = -(	an^{-1} 2 + 	an^{-1} 3)$,which is negative and cannot be $\frac{\pi}{4}$.If $x = \frac{1}{6}$,then $	an^{-1}(\frac{1}{3}) + 	an^{-1}(\frac{1}{2}) = 	an^{-1} \left( \frac{1/3 + 1/2}{1 - 1/6} ight) = 	an^{-1} \left( \frac{5/6}{5/6} ight) = 	an^{-1}(1) = \frac{\pi}{4}$.Thus,the only valid solution is $x = \frac{1}{6}$.

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

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