8.373. $\operatorname{tg} x+\operatorname{tg} \alpha+1=\operatorname{tg} x \operatorname{tg} \alpha$.

8.373. $\tan x+\tan \alpha+1=\tan x \tan \alpha$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos \alpha \neq 0 .\end{array}\right.$

Write the equation in the form

$\operatorname{tg} x+\operatorname{tg} \alpha=-(1-\operatorname{tg} x \operatorname{tg} \alpha) \Rightarrow \frac{\operatorname{tg} x+\operatorname{tg} \alpha}{1-\operatorname{tg} x \operatorname{tg} \alpha}=-1 \Rightarrow \operatorname{tg}(x+\alpha)=-1$,

$x+\alpha=-\frac{\pi}{4}+\pi k, x=-\alpha-\frac{\pi}{4}+\pi k, k \in Z$.

From the domain of definition, it follows that $\alpha \neq \frac{\pi}{4}$.

Answer: $x=-\alpha+\frac{\pi}{4}(4 k-1)$ for $\alpha \neq \frac{\pi}{4}, k \in Z$. For $\alpha=\frac{\pi}{4}$, there are no solutions.