4. The numbers $x$ and $y$ are such that the equalities $\operatorname{tg} x+\operatorname{tg} y=4$ and $3 \sin (2 x+2 y)=\sin 2 x \sin 2 y$ hold. Find $\operatorname{ctg} x \operatorname{ctg} y$.
Answer: $\frac{7}{6}$.

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

$$
\frac{\sin x \cos y+\sin y \cos x}{\cos x \cos y}=4 \Leftrightarrow \sin (x+y)=4 \cos x \cos y
$$

Transform the second equality:

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

Substitute $4 \cos x \cos y$ for $\sin (x+y)$ in the left part and transform the obtained equality (considering that $\cos x \cos y \neq 0$ due to the domain of definition):

$24 \cos x \cos y \cos (x+y)=4 \sin x \sin y \cos x \cos y \Leftrightarrow 6 \cos (x+y)=\sin x \sin y \Leftrightarrow$

$$
\Leftrightarrow 6 \cos x \cos y-6 \sin x \sin y=\sin x \sin y \Leftrightarrow 6 \cos x \cos y=7 \sin x \sin y
$$

From the last equality, it follows that $\sin x \sin y \neq 0$, so the expression $\operatorname{ctg} x \operatorname{ctg} y$ exists and $\operatorname{ctg} x \operatorname{ctg} y=\frac{7}{6}$.