Question

If $\sum _{n=1}^{20}\sum _{m=1}^{20}{\tan }^{-1}\left(\frac{m}{n}\right)=k\pi ,$ then the value of $k$ is

• A. $100$
• B. $200$
• C. $300$
• D. $400$

Answer 1

The correct option is A $100$

Let $S=\sum _{n=1}^{20}\sum _{m=1}^{20}{\tan }^{-1}\left(\frac{m}{n}\right)\cdots \left(1\right)$

Interchanging $m$ and $n$
$S=\sum _{m=1}^{20}\sum _{n=1}^{20}{\tan }^{-1}\left(\frac{n}{m}\right)\cdots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$
$2S=\sum _{m=1}^{20}\sum _{n=1}^{20}({\tan }^{-1}\left(\frac{m}{n}\right)+{\tan }^{-1}\left(\frac{n}{m}\right))2S=\sum _{m=1}^{20}\sum _{n=1}^{20}({\tan }^{-1}\left(\frac{m}{n}\right)+{\cot }^{-1}\left(\frac{m}{n}\right))$
$\Rightarrow 2S=\frac{\pi }{2}\times 20\times 20(\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2})\Rightarrow S=100\pi$