The set of values of x such that tan ^{-1}lef | vedclass

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(D) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Given equation: $	an ^{-1}\left(\frac{x}{x-2}ight

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$\left\{\frac{1}{3}, 4\right\}$

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(D) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Given equation: $	an ^{-1}\left(\frac{x}{x-2}ight)-	an ^{-1}\left(\frac{x}{2 x-1}ight)=	an ^{-1}\left(\frac{2}{3}ight)$.Applying the formula, we get:$	an ^{-1} \left( \frac{\frac{x}{x-2} - \frac{x}{2x-1}}{1 + \frac{x}{x-2} \cdot \frac{x}{2x-1}} ight) = 	an ^{-1} \left( \frac{2}{3} ight)$.Simplifying the expression inside the $	an ^{-1}$:$\frac{x(2x-1) - x(x-2)}{(x-2)(2x-1) + x^2} = \frac{2}{3}$.$\frac{2x^2 - x - x^2 + 2x}{2x^2 - x - 4x + 2 + x^2} = \frac{2}{3}$.$\frac{x^2 + x}{3x^2 - 5x + 2} = \frac{2}{3}$.Cross-multiplying:$3(x^2 + x) = 2(3x^2 - 5x + 2)$.$3x^2 + 3x = 6x^2 - 10x + 4$.$3x^2 - 13x + 4 = 0$.Factoring the quadratic equation:$(3x - 1)(x - 4) = 0$.Thus, $x = \frac{1}{3}$ or $x = 4$.Both values satisfy the domain of the original expression. Therefore, the set of values is $\left\{\frac{1}{3}, 4ight\}$.

The set of values of $x$ such that $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ is

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