4. The numbers $x$ and $y$ are such that the equalities $\operatorname{tg} x - \operatorname{tg} y = 7$ and $2 \sin (2 x - 2 y) = \sin 2 x \sin 2 y$ hold. Find $\operatorname{tg} x \operatorname{tg} y$.
Answer: $-\frac{7}{6}$.

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

$$
\frac{\sin x \cos y-\sin y \cos x}{\cos x \cos y}=3 \Leftrightarrow \sin (x-y)=7 \cos x \cos y
$$

Transform the second equality:

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

Substitute $7 \cos x \cos y$ for $\sin (x-y)$ in the left part and transform the obtained equality (considering that due to the domain of definition $\cos x \cos y \neq 0$):

$$
\begin{aligned}
& 7 \cos x \cos y \cos (x-y)=\sin x \sin y \cos x \cos y \Leftrightarrow 7 \cos (x-y)=\sin x \sin y \Leftrightarrow \\
& \quad \Leftrightarrow 7 \cos x \cos y+7 \sin x \sin y=\sin x \sin y \Leftrightarrow 7 \cos x \cos y=-6 \sin x \sin y \Leftrightarrow \operatorname{tg} x \operatorname{tg} y=-\frac{7}{6}
\end{aligned}
$$