11.2. On the board, there are 4 numbers. Vasya multiplied the first of these numbers by $\sin \alpha$, the second - by $\cos \alpha$, the third - by $\operatorname{tg} \alpha$, and the fourth - by $\operatorname{ctg} \alpha$ (for some angle $\alpha$) and obtained a set of the same 4 numbers (possibly in a different order). What is the maximum number of different numbers that could have been written on the board?
Answer: 3 numbers.

Solution. First, let's prove that at least one zero was written on the board. Indeed, suppose the numbers $a, b, c, d$ written on the board were non-zero, then their product $a b c d$ should equal the product of the new numbers $a b c d \sin \alpha \cos \alpha \operatorname{tg} \alpha \operatorname{ctg} \alpha$. But from $a b c d \neq 0$ it follows that $1=\sin \alpha \cos \alpha \operatorname{tg} \alpha \operatorname{ctg} \alpha=\frac{1}{2} \sin 2 \alpha$, which is impossible. Therefore, among the numbers on the board, there is a 0. Note that $\sin \alpha \cos \alpha \neq 0$ (and thus $|\sin \alpha| \neq 1,|\cos \alpha| \neq 1$), otherwise $\operatorname{tg} \alpha$ or $\operatorname{ctg} \alpha$ would be undefined. If among the remaining three numbers there is no zero, then similarly to the previous reasoning, the product of the three trigonometric functions equals 1. But $\cos \alpha \operatorname{tg} \alpha \operatorname{ctg} \alpha=\cos \alpha \neq 1, \quad \sin \alpha \operatorname{tg} \alpha \operatorname{ctg} \alpha=\sin \alpha \neq 1, \quad \sin \alpha \cos \alpha \operatorname{ctg} \alpha=\cos ^{2} \alpha \neq 1$, $\sin \alpha \cos \alpha \operatorname{tg} \alpha=\sin ^{2} \alpha \neq 1$. Therefore, there are at least two zeros on the board, meaning there are no more than 3 different numbers. Exactly three different numbers could have been, for example, in such a situation. The numbers $0,0,1$ and 2 were written on the board, and they were multiplied by $\sin \frac{\pi}{4}, \cos \frac{\pi}{4}, \operatorname{tg} \frac{\pi}{4}, \operatorname{ctg} \frac{\pi}{4}$ respectively.

Comment. Only proved that there is a $0-1$ point on the board.

Proved that there are no more than 3 different numbers on the board - 5 points.

Provided an example with 3 different numbers - 2 points.