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$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$
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$0$
$\begin{aligned} 2 \tan ^{-1} \frac{1}{3} &=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{1}{3} \times \frac{1}{3}}\right) \\ &=\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)=\tan ^{-1} \frac{3}{4} \\ \text { Now } 2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right) &=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=0 \end{aligned}$
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