${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{4}{33}\right)+\dots \mathrm{\infty }=$

$$\begin{gathered} {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{4}{33}\right)+........+\infty \\ We know that {\tan }^{-1}\left(A\right)-{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A-B}{1+AB}\right) \\ ={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(8\right)-{\tan }^{-1}\left(4\right)+.....+{\tan }^{-1}\left(2^{n-1}\right)-{\tan }^{-1}\left(2^{n-2}\right)+....+{\tan }^{-1}(\infty ) \\ ={\tan }^{-1}(\infty )-{\tan }^{-1}\left(1\right) \\ =\frac{\pi }{2}-\frac{\pi }{4} \\ =\frac{\pi }{4} \end{gathered}$$