Find $\arctan(x) + \arctan(y)$ in terms of $\arctan\left(\dfrac{x+y}{1-xy}\right)$.
I want to essentially prove this equation given in the textbook: $$ \arctan(x) + \arctan(y) = \begin{cases} \arctan\left(\dfrac{x+y}{1-xy}\right), &\text{if } xy\lt 1 \\[1.5ex] \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &\text{if } x\gt 0, y\lt 0\text{ and } xy \gt 1 \\[1.5ex] -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &\text{if } x\lt 0, y\lt 0\text{ and } xy \gt 1 \\ \end{cases}$$
By now i have proven the third case where $x<0$ and $y<0$ as follows: Let $A=\arctan(x)$ and $B=\arctan(y)$, and $A+b=\alpha$. If $-\pi<A+B<-\pi/2$, then $A<0,B<0 \iff x<0,y<0$. As $A+B<-\pi/2$ for the third case, thus: $$A<-(\pi/2+B) \\ \arctan(x)<-(\pi/2+\arctan(y)) \\ \because y<0, \qquad \therefore -(\pi/2+\arctan(y))=\arctan(1/y) \\ \therefore \arctan(x) < \arctan(1/y) \implies x<1/y \\ \therefore xy>1 $$ I am specifically not able to find such a condition in $x$ and $y$ for the first case, that is the case where $xy < 1$. How do i prove it using a similar approach to mine? (That is no differentiation, only trigonometry). I tried using the inequality $A+B\in(-\pi/2,\pi/2)$, but am stuck.
