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In the arctangent formula, we have that:


however, only for $uv<1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?

One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:


Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv<1$ would mean $\tan\alpha\tan\beta<1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2<\alpha+\beta<\pi/2$, i.e. that we have to stay in the same period of tangent.

However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation.

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up vote 4 down vote accepted

You are correct in that it is related to the period. Note however, that while the period is unimportant for the tan addition formula to hold, the arc tan functions are defined by restricting the range to$(\dfrac{-\pi}{2},\dfrac{\pi}{2})$ If $uv>1 $, $\arctan u +\arctan v$ is not in the range of the arctan function(principal branch). In that case $\arctan \dfrac{u+v}{1-uv}$ is not the sum of the arctan's, it is shifted by $\pi$, up or down. Note that $\tan(\arctan u +\arctan v)=\tan(\arctan \dfrac{u+v}{1-uv})$ regardless of $uv<1$.

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OK, I think I understand: since the result, $\arctan\left(\frac{u+v}{1-uv}\right)$ must be within $(-\pi/2,\pi/2)$, we obviously cannot have $\arctan{u}+\arctan{v}\notin(-\pi/2,\pi/2)$, and this is the case if $uv>1$. Is that correct? – Johnny Westerling Mar 10 '13 at 11:10
Yes that's about it, as far as I know. – Ishan Banerjee Mar 10 '13 at 11:13
Well, I guess that's a good explanation. Thank you! – Johnny Westerling Mar 10 '13 at 11:37

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 $-\frac\pi2\le \arctan x+\arctan y\le \frac\pi2, \arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$

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Very nice answer (+1) – Johnny Westerling Mar 11 '13 at 4:43
The last answer, at point (3), does not explain clearly why $$ \arctan x + \arctan y \in [-\frac{\pi}{2},\frac{\pi}{2}] $$ implies $xy \leqslant 1$. <br /><br /> I've published a rigorous proof of the sum of arctangents on this page. – user104981 Nov 2 '13 at 23:28
@MicheleDeStefano, The sole target of the answer is to find the value of $n$ for the different ranges of values of $\arctan x,\arctan y$. For $(3),$ the sum is already within the required range so $n=0$ – lab bhattacharjee Nov 5 '13 at 15:51
@lab bhattacharjee, Ok. I agree. But (3) does not answer the original question at the beginning of the thread. The demonstration I've posted does. – user106080 Nov 7 '13 at 7:15
Hye, sir. Probably can't remember me; but you helped me earlier in trigonometry. But may I again request you to please help in my latest quo? A comment is enough! Please forgive me if I bothered you. But would be grateful if you help:) – MAFIA36790 Mar 10 '15 at 12:54

protected by Community Dec 1 '13 at 17:59

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