Arctangent addition formula for any range of the arguments

Question

The usual arctangent addition formula works only when the arguments are in the proper range, namely:

$$\arctan x+\arctan y=\arctan \frac{x+y}{1-xy}, \quad |xy| <1$$

However for the other arguments we need to add some integer multiple of $\pi$. The details can be seen in these questions:

A question about the arctangent addition formula.

proving $\tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$

I wasn't sure if there's a way to avoid this multiple choice and just write a single formula valid for all the cases. And that's when I had a thought. Using the double angle formula for tangents, we have:

$$\arctan x=2 \arctan \frac{\sqrt{1+x^2}-1}{x}$$

This seems to work for any values of $x$, but what' especially important, we have:

$$\left|\frac{\sqrt{1+x^2}-1}{x} \right| \leq 1, \qquad x \in \mathbb{R}$$

Which means, we can use the usual arctangent addition formula for this:

$$\arctan x+\arctan y=2 \arctan \frac{f(x)+f(y)}{1-f(x)f(y)} \\ f(x)=\frac{\sqrt{1+x^2}-1}{x}=\frac{x}{\sqrt{1+x^2}+1}$$

If we divide by $2$, this could be called the "arithmetic mean of arctangents"..

My questions are:

is this formula valid for any $x,y \in \mathbb{R}$?
is there a simpler formula for the whole domain, which doesn't require us to add some multiple of $\pi$ to the result, depending on the signs and the values of $x,y$?

As an example, we can easily show that:

$$\arctan 2+ \arctan 3=2 \arctan (1+ \sqrt{2})=\frac{3 \pi}{4}$$

Not very reflexively, it seems to me that the jumps from $\tan (x)$ to $ ±\infty$ must be for something in your problem. — Piquito, Nov 01 '18 at 13:13
Note that, at least for $x > 0$, $f(x) = \exp \operatorname{arcsch} x$. — Travis Willse, Nov 01 '18 at 13:17

score 1 · Accepted Answer · answered Nov 04 '18 at 08:42

Your doubling formula essentially ensures that we are working in interval $(-\pi/4,\pi/4)$ instead of the usual $(-pi/2,\pi/2)$.

The addition formula $$\arctan x+\arctan y=\arctan z, z=\frac{x+y} {1-xy}$$ is true if the value of left side lies in $(-\pi/2,\pi/2)$. This should be obvious as by definition the right side lies in this interval. By using doubling formula you ensure that the arctan values lie in $(-\pi/4,\pi/4)$ and the sum of two such values lies in $(-\pi/2,\pi/2)$. Thus the resulting formula for addition of arctan values holds true for all the real values of $x, y$.

Arctangent addition formula for any range of the arguments

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