Eliminate $x$ and $y$ from the following system of equations:

$$
\begin{aligned}
\operatorname{tan} x+\operatorname{tan} y & =a \\
\operatorname{cot} x+\operatorname{cot} y & =b \\
x+y & =c
\end{aligned}
$$
From (2) and based on $\operatorname{ctg} \alpha=\frac{1}{\operatorname{tg} \alpha}$

$$
\frac{\operatorname{tg} x+\operatorname{tg} y}{\operatorname{tg} x \cdot \operatorname{tg} y}=b
$$

and considering (1)

$$
\operatorname{tg} x \cdot \operatorname{tg} y=\frac{a}{b}
$$

It is known that

$$
\operatorname{tg}(x+y)=\frac{\operatorname{tg} x+\operatorname{tg} y}{1-\operatorname{tg} x \cdot \operatorname{tg} y}
$$

Substituting the values of $x+y, \operatorname{tg} x+\operatorname{tg} y$ and $\operatorname{tg} x \cdot \operatorname{tg} y$ from (3), (1), and (4) into (5):

$$
\operatorname{tg} c=\frac{a}{1-\frac{a}{b}}=\frac{1}{\frac{1}{a}-\frac{1}{b}}
$$

or

$$
\operatorname{ctg} c=\frac{1}{a}-\frac{1}{b}
$$

Solved by: Durst E., Kántor S.