# Linear equation of the Tangent function

Let's say we have the following equation:

$\tan(x+y)=a(\tan(x)+\tan(y))$, where $a$ is a nonzero real number.

What can we say about $x$ and $y$? Is there a general solution?

• What is $tan(A+B)$? Jul 10, 2018 at 23:08
• @Rumplestillskin If you are referring to the formula, I thought the question would be simpler in terms of linearity and not product.
– Asix
Jul 10, 2018 at 23:10
• Expanding the left-hand side looks like it would simplify things tremendously since you then have a factor of $\tan x+\tan y$ on both sides, effectively leaving you only two cases to consider.
– amd
Jul 10, 2018 at 23:44

If $a=1$ then addition formula for tangent gives $\tan x \tan y=0,$ so one of the tanents must be $0,$ and it seems that's equivalent. But this $a=1$ case is a bit too simple. Still starting with the addition formula one might derive a relation between $x,y$ to characterize your relation.
Since $\tan(x+y) =\dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$, if $\tan(x+y)=a(\tan(x)+\tan(y))$ then $a = \dfrac1{1-\tan(x)\tan(y)}$.
Therefore, given $a$ and either of $x, y$, say $x$, $1-\tan(x)\tan(y) =\dfrac1{a}$ so $\tan(y) =\dfrac{1-1/a}{\tan(x)}$.
• Don’t you need to consider the case $\tan(x)+\tan(y)=0$ separately?