$$\sum_{n=1}^{10}\sum_{m=1}^{10} \tan^{-1}\left(\frac{m}{n}\right)=k\pi $$ Should I use $$\tan^{-1}{x_1}+\tan^{-1}{x_2}+\tan^{-1}{x_3}+\cdots \tan^{-1}{x_n} =\tan^{-1}\left({\frac{S_1 -S_3 +S_5 -S_7+ \cdots}{1- S_2 + S_4 - S_6 \cdots}}\right) $$ where $S_k$ denotes the sum of products of $x_1,x_2,x_3, \cdots , x_n$ taken $k$ at a time.
Or
Is there any easier method to evaluate it.
Following Neretin's suggestion we have that for any positive integer $N$: $$\begin{align*}\sum_{n=1}^{N}\sum_{m=1}^{N} \tan^{-1}\left(\frac{m}{n}\right)&=\sum_{n=1}^{N}\tan^{-1}\left(1\right)+\sum_{n=2}^{N}\sum_{m=1}^{n-1} \left(\tan^{-1}\left(\frac{m}{n}\right)+\tan^{-1}\left(\frac{n}{m}\right)\right)\\ &=\frac{\pi}{4}\sum_{n=1}^{N}1+\frac{\pi}{2}\sum_{n=2}^{N}\sum_{m=1}^{n-1} 1\\ &=N\cdot\frac{\pi}{4}+\frac{\pi}{2}\sum_{n=2}^{N}(n-1)\\ &=N\cdot\frac{\pi}{4}+\frac{N(N-1)}{2}\cdot\frac{\pi}{2}=\frac{N^2\pi}{4}. \end{align*}$$
P.S. $\tan^{-1}x+\tan^{-1}{1\over x}={\pi\over2}$ holds for $x>0$ and it can be easily verified by noting that the derivative of the l.h.s. is zero.