Condition for $\tan A\tan B=\tan C\tan D$ Here, it is claimed that $$\tan A\tan B=\tan C\tan D$$ if one of the four following conditions holds $$\displaystyle A\pm B=C\pm D$$
If it is true, how to prove this?
$\tan(x\pm y)$ did not help much.
I am expecting some relation among $A,B,C,D$ 
 A: $$\text{If }\tan A\tan B=\tan C\tan D,$$
$$\text{we have,}\frac{\sin A\sin B}{\cos A\cos B}=\frac{\sin C\sin D}{\cos C\cos D}$$
Applying Componendo and dividendo,
$$\frac{\cos A\cos B-\sin A\sin B}{\cos A\cos B+\sin A\sin B}=\frac{\cos C\cos D-\sin C\sin D}{\cos C\cos D+\sin C\sin D}$$
$$\iff \frac{\cos(A+B)}{\cos(A-B)}=\frac{\cos(C+D)}{\cos(C-D)}$$
$$\iff \cos(A+B)\cos(C-D)=\cos(C+D)\cos(A-B) $$
So, the condition meant by the link in the question seems to be
either $A+B=2n_1\pi\pm(C-D)$ and $A-B=2n_2\pi\pm(C+D)$
or $A-B=2n_3\pi\pm(C+D)$ and $A+B=2n_4\pi\pm(C-D)$  where $n_i$s are integers
as $\cos x=\cos y\implies x=2n\pi\pm y$ where $n$ is any integer
A: 
Above is the graph for $\tan(x)\tan(y)=2$. Clearly the relation is nothing like the one in the question. So $A\pm B=C\pm D$ is neither necessary or sufficient. I think that it's going to be hard to find any way of describing $\tan(A)\tan(B)=\tan(C)\tan(D)$ other than just "$\tan(A)\tan(B)=\tan(C)\tan(D)$".
You could try writing $\tan(\theta)$ as $\frac{x^2-1}{i(x^2+1)}$ where $x=e^{i\theta}$ and then rearranging, but it doesn't look like it's going to have a nicer form than the one it starts out in.
