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The value of $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{7}{8}$ is
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The correct answer is:
$\tan ^{-1} 15$
$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{7}{8}$
$=\left[\frac{\frac{1}{2}+\frac{1}{3}+\frac{7}{8}-\frac{1}{2} \times \frac{1}{3} \times \frac{7}{8}}{1-\frac{1}{2} \times \frac{1}{3}-\frac{1}{3} \times \frac{7}{8}-\frac{7}{8} \times \frac{1}{2}}\right]$
$\left[\because \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right.$
$\left.\quad=\tan ^{-1}\left(\frac{x+y+z-x y z}{1-x y-y z-z x}\right)\right]$
$=\tan ^{-1}\left[\frac{\frac{41}{24}-\frac{7}{48}}{1-\frac{1}{6}-\frac{7}{24}-\frac{7}{16}}\right]$
$=\tan ^{-1}\left[\frac{\frac{75}{48}}{1-\frac{43}{48}}\right]=\tan ^{-1}\left(\frac{75}{48-43}\right)$
$=\tan ^{-1}\left[\frac{75}{5}\right]=\tan ^{-1} 15$
$=\left[\frac{\frac{1}{2}+\frac{1}{3}+\frac{7}{8}-\frac{1}{2} \times \frac{1}{3} \times \frac{7}{8}}{1-\frac{1}{2} \times \frac{1}{3}-\frac{1}{3} \times \frac{7}{8}-\frac{7}{8} \times \frac{1}{2}}\right]$
$\left[\because \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right.$
$\left.\quad=\tan ^{-1}\left(\frac{x+y+z-x y z}{1-x y-y z-z x}\right)\right]$
$=\tan ^{-1}\left[\frac{\frac{41}{24}-\frac{7}{48}}{1-\frac{1}{6}-\frac{7}{24}-\frac{7}{16}}\right]$
$=\tan ^{-1}\left[\frac{\frac{75}{48}}{1-\frac{43}{48}}\right]=\tan ^{-1}\left(\frac{75}{48-43}\right)$
$=\tan ^{-1}\left[\frac{75}{5}\right]=\tan ^{-1} 15$
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