How to prove analytically $\frac{\pi}{4}-\arctan \frac{m}{n}=\arctan \frac{1}{m+n}$ applies to at least all rational numbers?

Question

Using multiple triangles to find the values of some arctan angles in a trigonometry question so that I could use an exact expression I wondered how could I prove that the following relation applies to all possible rational numbers without using geometry to prove individual case by case:

$\frac{\pi}{4}-\arctan \frac{m}{n}=\arctan \frac{1}{m+n}$

Where the original geometrical thought was

$\frac{\pi}{4}-\arctan \frac{1}{2}=\arctan \frac{1}{3}$

$\frac{\pi}{4}-\arctan \frac{2}{3}=\arctan \frac{1}{5}$

And lastly, could I use this more widely with complex numbers and if so with what limitations?

Answer 1

I'm not sure exactly what you are asking here; your equations are wrong. But if $0<m<n$ then $$\frac\pi4-\arctan\frac mn=\arctan\frac{n-m}{m+n}.$$ In particular, $$\frac\pi4-\arctan\frac m{m+1}=\arctan\frac{1}{2m+1}$$ so that $$\frac\pi4-\arctan\frac12=\arctan\frac{1}{3},$$ $$\frac\pi4-\arctan\frac23=\arctan\frac{1}{5},$$ $$\frac\pi4-\arctan\frac34=\arctan\frac{1}{7}$$ etc. Is this what you were aiming at?

Answer 2

Notice that $$\tan \left(\frac{\pi}{4} - \arctan(\frac{m}{n}) \right) = \frac{n-m}{n+m} $$