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Question Number 108802 by 150505R last updated on 19/Aug/20

Answered by mr W last updated on 19/Aug/20

$${\tan }^{-1}\left(\frac{i}{j}\right)+{\tan }^{-1}\left(\frac{j}{i}\right)=\frac{\pi }{2}$$$${\tan }^{-1}\left(\frac{i}{i}\right)=\frac{\pi }{4}$$$$\underset{i=1}{\sum ^n}\underset{j=1}{\sum ^n}{\tan }^{-1}\left(\frac{i}{j}\right)=\frac{n^2-n}{2}\times \frac{\pi }{2}+n\times \frac{\pi }{4}$$$$=\frac{n^2\pi }{4}=\frac{{20}^2\pi }{4}=100\pi$$

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