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Prove that:
$\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$
$\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$
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Verified Answer
$\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}$
$=\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \times \frac{1}{7}}\right)+\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \times \frac{1}{8}}\right)$
$=\tan ^{-1} \frac{6}{17}+\tan ^{-1} \frac{11}{23}=\tan ^{-1}\left(\frac{325}{325}\right)$
$=\tan ^{-1} 1=\frac{\pi}{4}$
$=\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \times \frac{1}{7}}\right)+\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \times \frac{1}{8}}\right)$
$=\tan ^{-1} \frac{6}{17}+\tan ^{-1} \frac{11}{23}=\tan ^{-1}\left(\frac{325}{325}\right)$
$=\tan ^{-1} 1=\frac{\pi}{4}$
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