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

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

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$\frac{1}{6}$

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(B) Given the equation: $	an ^{-1}(2 x)+	an ^{-1}(3 x)=\frac{\pi}{4}$ Using the identity $	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: $(6 x - 1)(x + 1) = 0$ This gives $x = \frac{1}{6}$ or $x = -1$. Since the problem states $x > 0$,we discard $x = -1$. Therefore,$x = \frac{1}{6}$.

If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$,where $x>0$,then $x=$

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