Let $\tan^{-1}\left(\tfrac{1}{3}\right)=\theta$ and $\tan^{-1}\left(\tfrac{1}{7}\right)=\phi$
$\enspace\therefore\quad\tan\theta=\tfrac{1}{3}$ and $\tan\phi=\tfrac{1}{7}$
Now $2\tan^{-1}\left(\tfrac{1}{3}\right)+$ $\tan^{-1}\left(\tfrac{1}{7}\right)=$ $2\theta+\phi$
and $\tan(2\theta+\phi)$
$=\dfrac{\tan2\theta+\tan\phi}{1-\tan2\phi\tan\phi}$
$=\dfrac{\dfrac{2\tan\theta}{1-\tan^{2}\theta}+\tan\phi}{1-\left(\dfrac{2\tan\theta}{1-\tan^{2}\theta}\right)\times\tan\phi}$
$=\dfrac{\dfrac{\tfrac{2}{3}}{1-\tfrac{1}{9}}+\tfrac{1}{7}}{1-\left(\dfrac{\tfrac{2}{3}}{1-\tfrac{1}{9}}\right)\times\tfrac{1}{7}}$
$=\left(\dfrac{\tfrac{3}{4}+\tfrac{1}{7}}{1-\tfrac{3}{4}\times\tfrac{1}{7}}\right)\times\tfrac{28}{28}$
$\enspace\therefore\quad2\theta+\phi=$ $\tfrac{\pi}{4}+k\pi$, for some $k\in\mathbb{Z}$.
But $\tan^{-1}\left(\tfrac{1}{3}\right)=\theta$ and $\tan^{-1}\left(\tfrac{1}{7}\right)=\phi$ both lie between $0$ and $\tfrac{\pi}{4}$.
$\enspace\therefore\quad2\theta+\phi=$
$\tfrac{\pi}{4}$
$\enspace\therefore\quad2\tan^{-1}\left(\tfrac{1}{3}\right)+\tan^{-1}\left(\tfrac{1}{7}\right)=$
$\tfrac{\pi}{4}$
