Condition for $\tan A\tan B=\tan C\tan D$

Question

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 $A,B,C,D$

Consider $A = 0$, $B = \pi/4$, $C = D = \pi/8$. Then the LHS is zero but the RHS is not. — , Sep 29 '13 at 09:17
@T.Bongers, clearly, $A= \frac\pi2,B=\frac\pi4, C=D=\frac\pi8$ do not satisfy the proposition. Have they assumed all the tangent ratios to be non-zero & finite? — lab bhattacharjee, Sep 29 '13 at 09:23
Take $A = 0.001$ and $B = \pi/4 - 0.001$ with the same $C$ and $D$, and it still doesn't work. — , Sep 29 '13 at 09:29
@T.Bongers, clearly, $A=0, C\pm D=\pm B\ne0$ or $A=\frac\pi2, C\pm D=\frac\pi2\pm A$ where $B,C\ne\pm\frac\pi2$ do not satisfy the proposition. — lab bhattacharjee, Sep 29 '13 at 09:30
@LeenDroogendijk, was going to add the condition to the question/ask him the point — lab bhattacharjee, Sep 29 '13 at 09:36
Never mind. It was a joke answer. He removed it, or somebody else removed it for him. Nice to see that you found the answer yourself. — Leen Droogendijk, Sep 30 '13 at 10:18

lab bhattacharjee · Answer 1 · 2013-09-29T10:44:06.327

$$\text{If }\tan A\tan B=\tan C\tan D,$$

$$\text{we have,}\frac{\sin A\sin B}{\cos A\cos B}=\frac{\sin C\sin D}{\cos C\cos D}$$

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

score 1 · Answer 2 · edited Jun 12 '20 at 10:38

tan(x)tan(y)=2

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.

Condition for $\tan A\tan B=\tan C\tan D$

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