10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$
Answer: 0.

First solution. From $\sin x = \operatorname{tg} y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$ or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$). Similarly, from the remaining equations, we obtain the inequalities $|\sin y| \geq |\sin z|$ and $|\sin z| \geq |\sin x|$. Thus, $|\sin x| \geq |\sin y| \geq |\sin z| \geq |\sin x|$. Therefore, $|\sin x| = |\sin y| = |\sin z|$. Since all inequalities have become equalities, we have $\sin x = \sin y = \sin z = 0$, and $\sin x + \sin y + \sin z = 0$.

Second solution. If one of the sines is zero, then the tangent equal to it is also zero, which means the sine in the numerator of the tangent is zero. Consequently, the other sines and tangents are also zero. In this case, $\sin x + \sin y + \sin z = 0$.

Suppose none of the sines are zero. Multiplying all three equations, we get $\sin x \sin y \sin z = \operatorname{tg} y \operatorname{tg} z \operatorname{tg} x = \frac{\sin x \sin y \sin z}{\cos x \cos y \cos z}$. Since $\sin x \sin y \sin z \neq 0$, we have $\cos x \cos y \cos z = 1$. This is only possible if $|\cos x| = |\cos y| = |\cos z| = 1$, which means the sines are zero, and the considered case is impossible.

Comment. A correct answer without justification - 0 points.

An answer obtained by considering an example - 1 point.

At least one case is incorrectly considered (or omitted) - no more than 3 points.