If $\arctan x=A,\arctan y=B;$ $\tan A=x,\tan B=y$
We know, $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
So, $$\tan(A+B)=\frac{x+y}{1-xy}$$
$$\implies\arctan\left(\frac{x+y}{1-xy}\right)=n\pi+A+B=n\pi+\arctan x+\arctan y $$ where $n$ is any integer
As the principal value of $\arctan z$ lies $\in[-\frac\pi2,\frac\pi2], -\pi\le\arctan x+\arctan y\le\pi$
$(1)$ If $\frac\pi2<\arctan x+\arctan y\le\pi, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y-\pi$ to keep $\arctan\left(\frac{x+y}{1-xy}\right)\in[-\frac\pi2,\frac\pi2]$
Observe that $\arctan x+\arctan y>\frac\pi2\implies \arctan x,\arctan y>0\implies x,y>0 $
$\implies\arctan x>\frac\pi2-\arctan y$
$\implies x>\tan\left(\frac\pi2-\arctan y\right)=\cot \arctan y=\cot\left(\text{arccot}\frac1y\right)\implies x>\frac1y\implies xy>1$
$(2)$ If $-\pi\le\arctan x+\arctan y<-\frac\pi2, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y+\pi$
Observe that $\arctan x+\arctan y<-\frac\pi2\implies \arctan x,\arctan y<0\implies x,y<0 $
Let $x=-X^2,y=-Y^2$
$\implies \arctan(-X^2)+\arctan(-Y^2)<-\frac\pi2$
$\implies \arctan(-X^2)<-\frac\pi2-\arctan(-Y^2)$
$\implies -X^2<\tan\left(-\frac\pi2-\arctan(-Y^2)\right)=\cot\arctan(-Y^2)=\cot\left(\text{arccot}\frac{-1}{Y^2}\right) $
$\implies -X^2<\frac1{-Y^2}\implies X^2>\frac1{Y^2}\implies X^2Y^2>1\implies xy>1 $
$(3)$ If $-\frac\pi2\le \arctan x+\arctan y\le \frac\pi2, \arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$
You are asking about a proof of the identity
$$
\tan\left(x\right) + \tan\left( y \right) = \frac{{\sin\left( {x + y} \right)}}{{\cos\left( x \right)\cos\left( y \right)}}
$$
Using $\tan(x)=\frac{\sin(x)}{\cos(x)}$, we get
$$\tan\left(x\right) + \tan\left( y \right) = \frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}\\
=\frac{\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)}{\cos(x)\cdot \cos(y)}$$
Using the identity $\sin(x+y)=\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)$ gives you the answer of your question.
Best Answer
To get rid of the discontinuities, multiply everything by $(1-\tan(A))\cos(A)$ which makes that we look for the zero's of function $$f(A)=(A+4) \sin (A)-(A+1) \cos (A)$$
As said in comments, the solutions will be closer and closer to $\left(n+\frac{1}{4}\right)\pi$. So, let $$A= \left(n+\frac{1}{4}\right)\pi+x$$ and expand to obtain $$6\cos(x)+(4x+(4n+1)\pi+10)\sin(x)=0$$
Now, using the very first terms of the Taylor expansions of the sine and cosine functions around $x=0$, you will have $$6+((4n+1)\pi+10) x+O(x^2)=0 \quad \implies x=-\frac 6{(4n+1)\pi+10 }$$ that is to say $$A_{(n)} \sim \left(n+\frac{1}{4}\right)\pi-\frac 6{(4n+1)\pi+10}$$