Prove that $2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}$
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Prove that $2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}$

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$\text { L.H.S }=  2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$

$=\tan ^{-1} \frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}+\tan ^{-1}\left(\frac{1}{7}\right) \quad\left[\therefore 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right]$

$=\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$

$=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \times \frac{1}{7}}\right)=\tan ^{-1}\left(\frac{21+4}{28-3}\right)$

$=\tan ^{-1}\left(\frac{25}{25}\right)=\tan ^{-1}(1)=\frac{\pi}{4}=\text { R.H.S. }$

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