We know that
$$\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex] \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex] -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
My question is: What is $\arctan(x) + \arctan(y)$ if $xy=1$
It is $\frac\pi2$ (assuming that $x,y>0$), because $\arctan(1)+\arctan(1)=\frac\pi2$ and$$\left(\arctan(x)+\arctan\left(\frac1x\right)\right)'=0.$$
If $x>0$, then $0<\arctan x<\pi/2$, and if $t=\arctan x$, then $\cot t=1/\tan t=1/x$, so that $\arctan(1/x)=\pi/2-t$. If $x>0$, then $\arctan x+\arctan (1/x)=\pi/2$.
As $\arctan$ is an odd function, if $x<0$, then $\arctan x+\arctan (1/x)=-\pi/2$.
