Problem 6. Calculate

$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2019 \pi}{47} \cdot \operatorname{tg} \frac{2020 \pi}{47}
$$
Answer: -2021.

Solution. From the formula

$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$

we express the product of tangents:

$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$

Then

$$
\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}=\frac{\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}}{\operatorname{tg}\left(\frac{(k+1) \pi}{47}-\frac{k \pi}{47}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-1
$$

Adding these equalities for all $k$ from 1 to 2019, we get that the expression in the condition equals

$$
\left(\operatorname{tg} \frac{2020 \pi}{47}-\operatorname{tg} \frac{\pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-2019
$$

Notice that

$$
\operatorname{tg} \frac{2020 \pi}{47}=\operatorname{tg}\left(43 \pi-\frac{\pi}{47}\right)=-\operatorname{tg} \frac{\pi}{47}
$$

which means $(*)$ equals -2021.