## Task A-3.1.

Solve the equation

$$
\operatorname{tg}(x+y)+\operatorname{ctg}(x-y)=\operatorname{tg}(x-y)+\operatorname{ctg}(x+y)
$$

and sketch in the plane the set of all its solutions.
## First Solution.

For $\operatorname{tg}(x+y)$ and $\operatorname{tg}(x-y)$ to be defined, it must not be the case that $x \pm y=\frac{\pi}{2}+k \pi$ for any $k \in \mathbb{Z}$. For $\operatorname{ctg}(x+y)$ and $\operatorname{ctg}(x-y)$ to be defined, it must not be the case that $x \pm y=k \pi$ for any $k \in \mathbb{Z}$.

From this, we have $x \pm y \neq \frac{k \pi}{2}$, for all $k \in \mathbb{Z}$, or equivalently, $x \neq \pm y+\frac{k \pi}{2}$, for all $k \in \mathbb{Z}$.

(1 point)

The given equation can be transformed as follows:

$$
\begin{aligned}
\operatorname{tg}(x+y)+\operatorname{ctg}(x-y) & =\operatorname{tg}(x-y)+\operatorname{ctg}(x+y) \\
\operatorname{ctg}(x-y)-\operatorname{tg}(x-y) & =\operatorname{ctg}(x+y)-\operatorname{tg}(x+y) \\
\frac{\cos (x-y)}{\sin (x-y)}-\frac{\sin (x-y)}{\cos (x-y)} & =\frac{\cos (x+y)}{\sin (x+y)}-\frac{\sin (x+y)}{\cos (x+y)} \\
\frac{\cos ^{2}(x-y)-\sin ^{2}(x-y)}{\sin (x-y) \cos (x-y)} & =\frac{\cos ^{2}(x+y)-\sin ^{2}(x+y)}{\sin (x+y) \cos (x+y)} \\
\frac{\cos (2 x-2 y)}{\frac{1}{2} \sin (2 x-2 y)} & =\frac{\cos (2 x+2 y)}{\frac{1}{2} \sin (2 x+2 y)} \\
\operatorname{ctg}(2 x-2 y) & =\operatorname{ctg}(2 x+2 y)
\end{aligned}
$$

We solve this equation as follows:

$$
2 x+2 y=2 x-2 y+m \pi, \quad \text { for } m \in \mathbb{Z}
$$

or equivalently, $y=\frac{m \pi}{4}, m \in \mathbb{Z}$.

The condition we determined at the beginning translates to $x \neq \pm \frac{m \pi}{4}+\frac{k \pi}{2}$,

or equivalently, $x \neq( \pm m+2 k) \frac{\pi}{4}, m, k \in \mathbb{Z}$.

(1 point)

(In other words, for even $m, x \neq \frac{l \pi}{2}(l \in \mathbb{Z})$; and for odd $m, x \neq \frac{\pi}{4}+\frac{l \pi}{2}(l \in \mathbb{Z})$.)

The solution set of the given equation is shown in the figure.

(2 points)