Finding an inverse trigonometric sum

Question

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.

Answer 1

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

Answer 2

Use the fact that

$$\arctan{\left ( \frac{p}{q} \right )} + \arctan{\left ( \frac{q}{p} \right )} = \frac{\pi}{2} $$

So all you need to do is arrange the sum to exploit the symmetry you need, i.e., $(p,q) \mapsto (q,p)$.

To do this without repeating, start on the bottom/left axes; there are $9$ non-diagonal points each, so the sum along there is $9 \pi/2$.

Then move one row up/column over to the right; we now have only $8$ (so we don't repeat). Keep moving up/right and get

$$(9+8+7+\cdots+1) \frac{\pi}{2} = 45 \frac{\pi}{2}$$

Now add in the diagonals, $10$ of them, contributing $\pi/4$ each (i.e., $\arctan{1}$), so the sum is

$$45 \frac{\pi}{2} + 10 \frac{\pi}{4} = 25 \pi$$