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

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

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$x=\frac{1}{6}$

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(D) Given the equation $	an ^{-1} 2 x+	an ^{-1} 3 x=\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$$6x^2 + 5x - 1 = 0$$(6x - 1)(x + 1) = 0$Thus,$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 not equal to $\frac{\pi}{4}$.Therefore,the only valid solution is $x = \frac{1}{6}$.

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

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