In the arctangent formula, we have that:

$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$

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:

$$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$

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.

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

• 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, 2013 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 $n=0$ Nov 5, 2013 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, 2013 at 7:15
• @Abcd, As $\arctan x<\dfrac\pi2,\arctan x+\arctan y$ can be $>\dfrac\pi2$ only if $\arctan y>0$ Feb 8, 2018 at 13:07

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 $uv<1$.

To explain why $\arctan(u) + \arctan(v) = \arctan(\frac{u+v}{1-uv})$ only if $uv < 1$, we need to recognise the obvious restriction of $$-\frac{\pi}{2} < \arctan(u) + \arctan(v) < \frac{\pi}{2}.$$

We'll consider three cases: $uv = 1$, $uv > 1$ and finally $uv < 1$.

If $uv = 1$ (the trivial case), $u$ and $v$ must have the same sign and $v = \frac{1}{u}$. From the identity $\arctan(u) + \arctan(\frac{1}{u}) = \pm\frac{\pi}{2}$:

$$\arctan(u) + \arctan(v) = \begin{cases} \frac{\pi}{2}, & \text{if u, v > 0}, \\[2ex] -\frac{\pi}{2}, & \text{if u, v < 0}. \\[2ex] \end{cases}$$

However, $\arctan(\frac{u+v}{1-uv})$ is obviously undefined when $uv=1$.

If $uv > 1$, both $u$ and $v$ must have the same sign again. Because of the behaviour of the inequality when dividing by a negative number, first consider the situation if $u, v > 0$: $$uv > 1 \implies v > \frac{1}{u},$$ $$\arctan(u) + \arctan(\frac{1}{u}) = \frac{\pi}{2},$$ $$\therefore \arctan(u) + \arctan(v) > \frac{\pi}{2}.$$

Similarly, if $u, v < 0$, $$uv > 1 \implies v < \frac{1}{u},$$ $$\arctan(u) + \arctan(\frac{1}{u}) = -\frac{\pi}{2},$$ $$\therefore \arctan(u) + \arctan(v) < -\frac{\pi}{2}.$$

Hence, we can see that whenever $uv > 1$, $\arctan(u) + \arctan(v) \notin (-\frac{\pi}{2}, \frac{\pi}{2})$.

If $uv < 1$, $u$ and $v$ may have different signs, but we can just consider $u > 0$, $u < 0$ and $u = 0$. If $u > 0$:

$$uv < 1 \implies v < \frac{1}{u},$$ $$\arctan(u) + \arctan(\frac{1}{u}) = \frac{\pi}{2},$$ $$\therefore \arctan(u) + \arctan(v) < \frac{\pi}{2}.$$

Similarly, if $u < 0$, $$uv < 1 \implies v > \frac{1}{u},$$ $$\arctan(u) + \arctan(\frac{1}{u}) = -\frac{\pi}{2},$$ $$\therefore \arctan(u) + \arctan(v) > -\frac{\pi}{2}.$$

If $u = 0$, consider the original equation $\arctan(u) + \arctan(v) = \arctan(\frac{u+v}{1-uv})$:

\begin{align} LHS &= \arctan(0) + \arctan(v) \\ &= \arctan(v) \\ RHS &= \arctan(\frac{0+v}{1-0}) \\ &= \arctan(v) \\ &= LHS \end{align}

To summarise, we have shown that $$-\frac{\pi}{2} < \arctan(u) + \arctan(v) < \frac{\pi}{2}$$ is only true if $uv < 1$.

Bonus: To extend this to the arctangent subtraction formula, $$\arctan(u) - \arctan(v) = \arctan(u) + \arctan(-v)$$ $$\therefore u(-v) < 1 \implies uv > -1.$$

What is $$\arctan{s}+\arctan{t}$$? It is the addition of two angles: $$\alpha+\beta$$

Starting from the image we first calculate $$h$$:

$$\frac{h}{s\times t}=s+t+h$$

Therefore:

$$h=s\times t\frac{s+t}{1-s\times t}$$

And:

$$\tan{(\alpha+\beta)}=\frac{h}{s\times t}$$

$$\alpha+\beta=\arctan{s}+\arctan{t}=\arctan{\frac{s+t}{1-s\times t}}$$

• I'm not sure how this answers the question which was asked: i.e. why do we need the product $st < 1$? Is there a way to see this from the diagram? Aug 21 at 13:48

A simpler explanation of why the identity $$\arctan u+\arctan v=\arctan\dfrac{u+v}{1-uv}$$ is valid if & only if $$uv <1$$.

By definition of the arctan function, as both sides have the same tangent, we only need to check under which condition $$-\frac\pi 2<\arctan u+\arctan v < \frac\pi2.$$ Let $$\alpha=\arctan u$$, $$\beta=\arctan v$$. A priori, $$\alpha+\beta\in(-\pi,\pi)$$, and on this interval, $$\alpha+\beta \in\Bigl(-\frac\pi 2,\frac\pi 2\Bigr)\iff \cos(\alpha+\beta)>0$$ Now $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta = \cos\alpha\cos\beta(1-\tan\alpha\tan\beta)\\=\underbrace{\cos\alpha\cos\beta}_{>0}\,(1-uv)$$ has the sign of $$1-uv$$.