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The value of $\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ is
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The correct answer is:
$\frac{\pi}{4}$
$\begin{aligned} & \tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2} \cdot \frac{1}{5}}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{7}{9}\right) \\ & =\tan ^{-1}\left(\frac{\frac{1}{8}+\frac{7}{9}}{1-\frac{1}{8} \cdot \frac{7}{9}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{72}{65}}{\frac{72}{9}}\right) \\ & =\tan ^{-1}(1) \\ & =\frac{\pi}{4}\end{aligned}$
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