4. The numbers $x$ and $y$ are such that the equalities $\operatorname{ctg} x - \operatorname{ctg} y = 2$ and $5 \sin (2 x - 2 y) = \sin 2 x \sin 2 y$ hold. Find $\operatorname{tg} x \operatorname{tg} y$.
Answer: $-\frac{6}{5}$.

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

$$
\frac{\cos x \sin y-\cos y \sin x}{\sin x \sin y}=3 \Leftrightarrow-\sin (x-y)=2 \sin x \sin y
$$

Transform the second equality:

$$
10 \sin (x-y) \cos (x-y)=4 \sin x \sin y \cos x \cos y
$$

Substitute $-2 \sin x \sin y$ 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):

$$
\begin{aligned}
-20 \sin x \sin y \cos (x-y) & =4 \sin x \sin y \cos x \cos y \Leftrightarrow-5 \cos (x-y)=\cos x \cos y \Leftrightarrow \\
\Leftrightarrow 5 \cos x \cos y+5 \sin x \sin y & =-\cos x \cos y \Leftrightarrow 6 \cos x \cos y=-5 \sin x \sin y
\end{aligned}
$$

From the last equality, it follows that $\cos x \cos y \neq 0$, so the expression $\operatorname{tg} x \operatorname{tg} y$ exists and $\operatorname{tg} x \operatorname{tg} y=-\frac{6}{5}$.