Solve the system of equations

$$
\begin{aligned}
x+y & =a \\
\operatorname{tan} x \cdot \operatorname{tan} y & =b
\end{aligned}
$$

where $a$ and $b$ are real parameters.
Solution. Since

$$
\frac{\cos (x-y)-\cos (x+y)}{2}=\sin x \cdot \sin y \quad \text { and } \quad \frac{\cos (x-y)+\cos (x+y)}{2}=\cos x \cdot \cos y
$$

the ratio of these is

$$
\frac{\sin x}{\cos x} \cdot \frac{\sin y}{\cos y}=\tan x \cdot \tan y=\frac{\cos (x-y)-\cos (x+y)}{\cos (x-y)+\cos (x+y)}=\frac{\cos (x-y)-\cos a}{\cos (x-y)+\cos a}=b
$$

(due to the tangent, $x \neq \frac{\pi}{2}+k \pi$ and $y \neq \frac{\pi}{2}+k \pi$, i.e., $\cos x \cdot \cos y \neq 0$). Therefore,

$$
\cos (x-y)-\cos a=b \cdot \cos (x-y)+b \cdot \cos a
$$

so $(1-b) \cdot \cos (x-y)=(1+b) \cdot \cos a$, and from this

$$
\cos (x-y)=\frac{1+b}{1-b} \cdot \cos a
$$

Hence,

$$
x-y= \pm\left(\arccos \left(\frac{1+b}{1-b} \cdot \cos a\right)\right)+2 k \cdot \pi \quad(k \in \mathbb{Z})
$$

Thus,

$$
\begin{aligned}
& x=\frac{(x+y)+(x-y)}{2}=\frac{a+2 k \cdot \pi \pm\left(\arccos \left(\frac{1+b}{1-b} \cdot \cos a\right)\right)}{2} \text { and } \\
& y=\frac{(x+y)-(x-y)}{2}=\frac{a-2 k \cdot \pi \mp\left(\arccos \left(\frac{1+b}{1-b} \cdot \cos a\right)\right)}{2}(k \in \mathbb{Z})
\end{aligned}
$$

If $b=1$, then $\cos (x-y)-\cos a=\cos (x-y)+\cos a$, so $0=2 \cdot \cos a$, i.e., $0=\cos (x+y)$. Therefore, if $a=\frac{\pi}{2}+k \pi$, there are infinitely many solutions $\left(y=\frac{\pi}{2}+k \pi-x\right)$, otherwise there are no solutions.