The number of solutions of tan^{-1} 2x + tan^ | vedclass

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(B) 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}

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$1$

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(B) 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)$ for $AB $	an^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} 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:$6x^2 + 6x - x - 1 = 0$.$6x(x + 1) - 1(x + 1) = 0$.$(6x - 1)(x + 1) = 0$.So,$x = \frac{1}{6}$ or $x = -1$.Check the solutions in the original equation:For $x = \frac{1}{6}$,$	an^{-1} (2/6) + 	an^{-1} (3/6) = 	an^{-1} (1/3) + 	an^{-1} (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}$. This is a valid solution.For $x = -1$,$	an^{-1} (-2) + 	an^{-1} (-3) = -(	an^{-1} 2 + 	an^{-1} 3)$,which is negative and cannot be $\frac{\pi}{4}$.Thus,there is only $1$ solution.

The number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

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