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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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Verified Answer
The correct answer is:
$\left\{\frac{1}{3}, 4\right\}$
We have,
$\begin{aligned}
& \tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right) \\
& \tan ^{-1} \frac{\left(\frac{x}{x-2}-\frac{x}{2 x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2 x-1}\right)}=\tan ^{-1} \frac{2}{3} \\
& \Rightarrow \frac{2 x^2-x-x^2+2 x}{2 x^2-4 x-x+2+x^2}=\frac{2}{3} \Rightarrow \frac{x^2+x}{3 x^2-5 x+2}=\frac{2}{3} \\
& \Rightarrow \quad 3 x^2+3 x=6 x^2-10 x+4 \\
& \Rightarrow \quad 3 x^2-13 x+4=0 \Rightarrow(3 x-1)(x-4)=0 \\
& \Rightarrow \quad x=\frac{1}{3}, 4, \quad x \in\left\{\frac{1}{3}, 4\right\}
\end{aligned}$
$\begin{aligned}
& \tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right) \\
& \tan ^{-1} \frac{\left(\frac{x}{x-2}-\frac{x}{2 x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2 x-1}\right)}=\tan ^{-1} \frac{2}{3} \\
& \Rightarrow \frac{2 x^2-x-x^2+2 x}{2 x^2-4 x-x+2+x^2}=\frac{2}{3} \Rightarrow \frac{x^2+x}{3 x^2-5 x+2}=\frac{2}{3} \\
& \Rightarrow \quad 3 x^2+3 x=6 x^2-10 x+4 \\
& \Rightarrow \quad 3 x^2-13 x+4=0 \Rightarrow(3 x-1)(x-4)=0 \\
& \Rightarrow \quad x=\frac{1}{3}, 4, \quad x \in\left\{\frac{1}{3}, 4\right\}
\end{aligned}$
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