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$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to
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The correct answer is:
$\frac{\pi}{4}$
$\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{9}\right)}\right) +\tan ^{-1}\left(\frac{1}{7}\right)\\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\ & =\tan ^{-1}\left(\frac{\left(\frac{3}{4}\right)+\left(\frac{1}{7}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{1}{7}\right)}\right)=\tan ^{-1}\left(\frac{25}{25}\right)=\frac{\pi}{4}\end{aligned}$
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