If the sum of  $\sum _{n=1}^{10}$$\sum _{m=1}^{10}$${\tan }^{-1}\left(\frac{m}{n}\right)=k\pi$then the value of k is

${\tan }^{-1}\left(\frac{1}{1}\right)+.{\tan }^{-1}\left(\frac{1}{2}\right)..................+{\tan }^{-1}\left(\frac{1}{10}\right)$

$+{\tan }^{-1}\left(\frac{2}{1}\right)++{\tan }^{-1}\left(\frac{2}{2}\right)\mathrm{...................}+{\tan }^{-1}\left(\frac{2}{10}\right)$

$$\begin{gathered} +\mathrm{..............} \\ +{\tan }^{-1}\left(\frac{10}{1}\right)+\mathrm{...................}+{\tan }^{-1}\left(\frac{10}{10}\right) \end{gathered}$$

$\begin{array}{l}=10\times \frac{\pi }{4}+45\times \frac{\pi }{2}\\ =25\pi \end{array}$$\left(\because {\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{2}{1}=\frac{\mathrm{\pi }}{2}, and so on\right)$