TITLE: Finding an inverse trigonometric sum
QUESTION [8 upvotes]: How do I prove that the following equality holds- $$\sum_{p=1}^{10} \sum_{q=1}^{10} \arctan \left(\dfrac{p}{q}\right)=25\pi$$
I tried to create telescoping terms by using the $\arctan{A}-\arctan{B}$ formula, but it doesn't seem to be working out. Hints in the right direction and answers appreciated. Thank you.

REPLY [4 votes]: Using Simplifying an Arctan equation, 
$$\arctan \frac{a}{b}+\arctan \frac{b}{a}=\frac\pi2$$  for $\dfrac ab>0$
We have such $\dfrac{10\cdot10}2$ pairs