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

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(A) 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}i

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$\frac{1}{6}$

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(A) 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{2 x+3 x}{1-(2 x)(3 x)}ight)=\frac{\pi}{4}$ $	an ^{-1}\left(\frac{5 x}{1-6 x^2}ight)=\frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{5 x}{1-6 x^2}=	an\left(\frac{\pi}{4}ight)=1$ $5 x=1-6 x^2$ $6 x^2+5 x-1=0$ Factoring the quadratic equation: $(6 x-1)(x+1)=0$ So,$x=\frac{1}{6}$ or $x=-1$. Checking the values: If $x=-1$,$	an ^{-1}(-2)+	an ^{-1}(-3)$ is negative,which cannot be $\frac{\pi}{4}$. Thus,$x=\frac{1}{6}$ is the only valid solution.

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

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