Which angles satisfy the following system of equations:

$$
\begin{aligned}
x - y & = \frac{\pi}{3} \ldots \\
\operatorname{tg} x - \operatorname{tg} y & = \sqrt{3}
\end{aligned}
$$
The equation can be written as

$$
\frac{\sin (x-y)}{\cos x \cos y}=\sqrt{3}
$$

Since

$$
\sin (x-y)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}
$$

it follows that:

$$
2 \cos x \cos y=\cos (x+y)+\cos (x-y)=1
$$

However,

$$
\cos (x-y)=\cos \frac{\pi}{3}=\frac{1}{2}
$$

thus

$$
\cos (x+y)=\frac{1}{2} \text {, i.e., } x+y= \pm \frac{\pi}{3}+2 k \pi
$$

Therefore, we have two systems of equations:
I. $x-y=\frac{\pi}{3}$ and $x+y=\frac{\pi}{3}+2 k \pi$. From this, $x=\frac{\pi}{3}+k \pi$ and $y=k \pi$, where $k$ is any positive or negative integer, or zero.

![](https://cdn.mathpix.com/cropped/2024_05_02_a7b0e313c71aeedea173g-1.jpg?height=81&width=1044&top_left_y=889&top_left_x=210)

Jenő Tornai (Kegyesrendi g. VI. o. Veszprém).