To calculate $\operatorname{tg} x+\operatorname{tg} y$, given that
(1) $\sin 2 x+\sin 2 y=a$,
(2) $\quad \cos 2 x+\cos 2 y=b$.
Let's introduce (for the sake of a more convenient description) the variables

$$
\alpha=x+y, \quad \beta=x-y
$$

The pair of variables $\alpha, \beta$ and the pair $x, y$ clearly determine each other uniquely. Substitute our new variables into equations (1) and (2). On the left side of (1), $\sin 2 x+\sin 2 y=\sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin \alpha \cdot \cos \beta$ stands, so (3) along with (1) is equivalent to the equation

$$
2 \sin \alpha \cos \beta=a
$$

Similarly, we get the equivalent of (2) as

$$
2 \cos \alpha \cos \beta=b
$$

Let's square these two equations and add their squares:

$$
4 \cos ^{2} \beta=a^{2}+b^{2}
$$

Therefore, the system of equations (1)-(2) is only solvable if

$$
a^{2}+b^{2} \leq 4
$$

If $a^{2}+b^{2}=0$, then $a=b=0$, and from (4) we have $\cos \beta=0$, which by itself implies the satisfaction of (1.a) and (2.a), so in this case the value of $\alpha$ is arbitrary. If $a^{2}+b^{2}>0$, then from (4) and (2.a) we have

$$
\cos \beta=\frac{1}{2} \sqrt{a^{2}+b^{2}} ; \quad \cos \alpha=\frac{b}{\sqrt{a^{2}+b^{2}}}
$$

These values can indeed be the cosines of some angles, and the angles $\alpha, \beta$ determined from these are solutions to the system of equations (1.a)-(2.a). Therefore, (5) is the necessary and sufficient condition for there to be a pair of angles $x, y$ for which (1) and (2) hold, and thus (5) must be satisfied to determine the value of

$$
t=\operatorname{tg} x+\operatorname{tg} y
$$

With our new variables, the value of $t$ (assuming it is defined, i.e., $x$ and $y$ are angles for which the tangent is defined) is:

$$
\begin{aligned}
t & =\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}=\frac{\sin x \cos y+\cos x \sin y}{\cos x \cos y}= \\
& =\frac{2 \sin (x+y)}{\cos (x+y)+\cos (x-y)}=\frac{2 \sin \alpha}{\cos \alpha+\cos \beta}
\end{aligned}
$$

We have seen that if $a=b=0$, then the value of $\cos \alpha$ is arbitrary, so in this case (1) and (2) do not uniquely determine the value of $t$. Since in this case $t=2 \operatorname{tg} \alpha$, the value of $t$ can be any real number, and it is also possible that $t$ is not defined.

If $a^{2}+b^{2}>0$, then from (4) we have $\cos \beta \neq 0$, so we can multiply the fraction for $t$ by $\cos \beta$:

$$
t=\frac{2 \sin \alpha \cos \beta}{\cos \alpha \cos \beta+\cos ^{2} \beta}
$$

The expression we are examining is defined if and only if the denominator is not zero. This condition, based on (4) and (2.a), is equivalent to the condition

$$
a^{2}+b^{2}+2 b \neq 0
$$

If this condition is satisfied, then

$$
t=\frac{4 a}{a^{2}+b^{2}+2 b}
$$

Therefore, (1) and (2) uniquely determine the value of $\operatorname{tg} x+\operatorname{tg} y$ if and only if $0<a^{2}+b^{2} \leq 4$, and (6) holds, and if these conditions are met, then in the case where (1) and (2) hold, $\operatorname{tg} x+\operatorname{tg} y$ is equal to the expression on the right side of (7).