atan(x)+atan(y)=atan((x+y)/(1-xy))
x,y=?
\(atan(x)+atany=atan(\frac{x+y}{1-xy})\\ let\;\;x=tanp\;\;\;and\;\;y=tanq\\ atan(tanp)+atan(tanq)=atan(\frac{tanp+tanq}{1-tanptanq})\\ p+q=atan(tan(p+q))\\ p+q=p+q\\ \)
There are restrictions on the domain and range but what I have discovered is that, allowing for the restrictions, this is an identity. (it is always true)
Determining the restricions is more confusing.
An obvious restricion is that 1-xy cannot be 0
\(xy\ne1\)
I can't get my head around the rest of the restrictions but here is what Wolfram alpha has to say about it..
https://www.wolframalpha.com/input/?i=atan(x)%2Batan(y)%3Datan((x%2By)%2F(1-xy))
What Melody explained means x and y could be anything unless x times y = 1.
Example x = .5 and y = 2 is impossible because the right hand side will be undefined. That will become atan of an undefined number which makes it more complicated. Actually I think this thing I already complicated enough XD