Are there numbers $x, y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$? My question is related to the following one: Is $\{\tan(x) : x\in \mathbb{Q}\}$ a group under addition?
It was shown that the above set is not closed under addition using the Lindemann-Weierstrass theorem. That tells us that $\tan(x) + \tan(y)$ need not be of the form $\tan(z)$ where $x,y,z \in \mathbb{Q}$.
An initial attempt in trying to show that the above set is not closed involved trying to find $x, y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$, because $\tan(x)$ is irrational when $x$ is a non-zero rational. However, I have not managed to find any rational numbers that do satisfy this property. I'm looking for non-trivial examples, so I'm discounting the case $x=-y$ for which $\tan(x) + \tan(y) = 0$.
The primary difficulty that I faced was that the decimal expansion gave me no clues whether I was looking at a rational number or an irrational number. I also tried playing with the formula for $\tan(x+y)$ but I was not able to derive anything from that either. I checked a few of my guesses on Wolfram|Alpha and it says that they are all transcendental.
I'm beginning to suspect that $\tan(x) + \tan(y)$ is never rational when $x$ and $y$ are rational, but I don't know how to prove that either. Does anyone have any ideas on the best way to proceed? Thank you for your help.

Let $x=\dfrac{p}{q}$ and $y=\dfrac{r}{s}$ and $z=\dfrac{1}{qs}$.
We observe that tan x and tan y are polynomials in tan z; further tan x + tan y is a non-constant polynomial if $x+y \neq 0$. Here some calculation should show that tan $A\theta$ + tan $B\theta$ is a non-constant polynomial for $A + B \neq 0$.
Hence we have:if tan x + tan y is rational, then tan z is the root of a polynomial with rational coefficients, contradicting the Generalized Linedmann Theorem, as pointed out by the OP.
