Prove that $\tan ^{-1}\left(\frac{2}{5}\right)+\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{12}\right)=\frac{\pi}{4}$
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Prove that $\tan ^{-1}\left(\frac{2}{5}\right)+\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{12}\right)=\frac{\pi}{4}$

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L.H.S. $=\tan ^{-1}\left(\frac{2}{5}\right)+\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{12}\right)$

$=\tan ^{-1}\left(\frac{\frac{2}{5}+\frac{1}{3}}{1-\frac{2}{5} \times \frac{1}{3}}\right)+\tan ^{-1}\left(\frac{1}{12}\right)$

$=\tan ^{-1}\left(\frac{11}{13}\right)+\tan ^{-1}\left(\frac{1}{12}\right)$

$=\tan ^{-1}\left(\frac{\frac{11}{13}+\frac{1}{12}}{1-\frac{11}{13} \times \frac{1}{12}}\right)$

$=\tan ^{-1}\left(\frac{132+13}{156-11}\right)=\tan ^{-1}\left(\frac{145}{145}\right)$

$=\tan ^{-1}(1)=\frac{\pi}{4}=\text { R.H.S. }$

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