Suppose we are to find
\[tan^{-1}\frac{1}{7}+2 tan^{-1}(\frac{1}{3})\]
.Let
\[\theta = tan^{-1}(\frac{1}{7}), \: \alpha = tan^{-1}(\frac{1}{3})\]
.Then
\[tan \theta = \frac{1}{7}, \: tan \alpha = \frac{1}{3}\]
.Now use
\[tan 2 \alpha = \frac{2tan \alpha}{1-tan^2 \alpha}\]
and \[tan)A+B)=\frac{tanA+tanB}{1-tanAtanB}\]
with \[A= \theta, B=2 \theta\]
.\[\begin{equation} \begin{aligned} tan(\theta+2 \alpha) & =\frac{tan \theta+tan 2 \alpha}{1-tan \theta tan 2 \alpha} \\ &= \frac{tan \theta +\frac{2tan \theta}{1-tan^2 \theta}}{1-tan \theta \frac{2tan \alpha}{1-tan^2 \alpha}} \\ &= \frac{1/7-\frac{2(1/3)}{1-(1/3)^2}}{1-(1/7)\frac{2(1/3)}{1-(1/3)^2}} \\ &= \frac{1-3/28}{1-1/7(3/4)} \\ &= (25/28)/(25/28)=1 \end{aligned} \end{equation} \]
Then
\[\theta + 2 \alpha = tan^{-1}(1)= \frac{\pi}{4}\]