4. The numbers $x$ and $y$ are such that the equalities $\operatorname{ctg} x+\operatorname{ctg} y=3$ and $2 \sin (2 x+2 y)=\sin 2 x \sin 2 y$ hold. Find $\operatorname{ctg} x \operatorname{ctg} y$.
Answer: $\frac{3}{2}$.

Solution. On the domain of definition, the first equality is equivalent to the following:

$$
\frac{\cos x \sin y+\cos y \sin x}{\sin x \sin y}=3 \Leftrightarrow \sin (x+y)=3 \sin x \sin y
$$

Transform the second equality:

$$
4 \sin (x+y) \cos (x+y)=4 \sin x \sin y \cos x \cos y
$$

Substitute $3 \sin x \sin y$ for $\sin (x+y)$ in the left part and transform the obtained equality (considering that $\sin x \sin y \neq 0$ due to the domain of definition):

$$
\begin{aligned}
& 12 \sin x \sin y \cos (x+y)=4 \sin x \sin y \cos x \cos y \Leftrightarrow 3 \cos (x+y)=\cos x \cos y \Leftrightarrow \\
& \quad \Leftrightarrow 3 \cos x \cos y-3 \sin x \sin y=\cos x \cos y \Leftrightarrow 2 \cos x \cos y=3 \sin x \sin y \Leftrightarrow \operatorname{ctg} x \operatorname{ctg} y=\frac{3}{2}
\end{aligned}
$$