inverse trigonometry ranges

Question

$$\arctan(x)+\arctan(y)=a+b\left(\arctan\frac{x+y}{1-xy}\right)$$

Find sum of all distinct possible values of $a$ and $b$. Also when we plot this graph and keep on changing the values of $a$ and $b$ we get different curves for different values.Is there any deeper insight for that?

Your question requires some refinements - the sum to find should allow what - a solution for x and y? A graph for xy? — Moti, Jan 16 '21 at 17:57
https://math.stackexchange.com/questions/326334/a-question-about-the-arctangent-addition-formula — lab bhattacharjee, Jan 16 '21 at 18:23
The requirement is to find values of a and b so that the given relation holds true . Then the sum of the distinct values of those obtained . — Soura Deep, Jan 16 '21 at 18:27
Actually , the answer is 1 because by def. We have if xy<1, a= 0,b=1 — Soura Deep, Jan 16 '21 at 18:28
If xy>1, x,y<0 a=-π ,b=1 but I was wondering if there is any other rigorous way to solve this — Soura Deep, Jan 16 '21 at 18:29
Do you mean $$\arctan(x)+\arctan(y)=a+b\left(\arctan{\frac{x+y}{1-xy}}\right)$$? — A-Level Student, Jan 16 '21 at 19:07

score 0 · Answer 1 · answered Jan 16 '21 at 19:45

If you do mean $$\arctan(x)+\arctan(y)=a+b\left(\arctan\frac{x+y}{1-xy}\right)$$ then this is easily solved. For $\lvert xy\rvert<1$, we have $$\arctan(x)+\arctan(y)=\arctan\frac{x+y}{1-xy}$$ However, if $\lvert xy\rvert\not<1 $ then either $$\arctan(x)+\arctan(y)=\pi+\arctan\frac{x+y}{1-xy}$$ or $$\arctan(x)+\arctan(y)=-\pi+\arctan\frac{x+y}{1-xy}$$ Can you solve it now?

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inverse trigonometry ranges

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