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As mentioned here it is known that one can write $\operatorname{Arctg}(a) + \operatorname{Arctg}(b)$ as a form of $\operatorname{Arctg}\frac{a+b}{1- a\: b}$ plus a constant in the case where $a \:b<1$ or $a \:b>1$. I am wondering what happens when $a b=1$. Is there a formula for this case?

Bernard
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Marco
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2 Answers2

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It is well known that $$\arctan x+\arctan\frac1x= \begin{cases}\phantom{-}\dfrac\pi 2&\text{if } x>0,\\[0.5ex]-\dfrac\pi 2 &\text{if } x<0.\end{cases}$$

Bernard
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  • I am not sure, because one can, for example, consider $a= \frac{p+c}{t}$ and $b=\frac{t}{p+c}$, if $p=-c$ the answer to my question is not clear – Marco Nov 18 '19 at 09:28
  • This is the case $x=0$, which I do not consider ($\frac1x$ is not defined). – Bernard Nov 18 '19 at 10:18
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Hint:

$ab = 1 \rightarrow a = b = 1$, so

$$\arctan (1) + \arctan (1) = \dfrac {1+1}{1-1} \rightarrow \dfrac 2 0 $$

Since $\dfrac 2 0$ is undefined, then

$$\arctan \dfrac 2 0 = \dfrac \pi 2.$$

Also, since $\arctan 1 = \dfrac {\pi}{4}$,

$$\arctan 1 + \arctan 1 = \dfrac \pi 4 + \dfrac \pi 4 = \dfrac \pi 2.$$

In other words, you can define the behavior of $\arctan (a) + \arctan (b)$ as follows:

$$\arctan (a)+\arctan(b)= \begin{cases}\phantom{-}\dfrac{a+b}{1-ab}&\text{if } a \neq b,\\[0.5ex]\phantom {-}\dfrac \pi 2 &\text{if } a=b.\end{cases}$$

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