Problem 1182

13. Let $u_{k}=\tan \left[\frac{\pi}{3}\left(1+\frac{3^{k}}{3^{n}-1}\right)\right], v_{k}=\tan \left[\frac{\pi}{3}\left(1-\frac{3^{k}}{3^{n}-1}\right)\right], t_{k}=\tan \frac{3^{k-1} \pi}{3^{n}-1}$
$$u_{k}=\tan \left(\frac{\pi}{3}+\frac{3^{k-1} \pi}{3^{n}-1}\right)=\frac{\sqrt{3}+t_{k}}{1-\sqrt{3} t_{k}}, v_{k}=\tan \left(\frac{\pi}{3}-\frac{3^{k-1} \pi}{3^{n}-1}\right)=\frac{\sqrt{3}-t_{k}}{1+\sqrt{3} t_{k}}$$

Since $\tan 3 \alpha=\frac{3 \tan \alpha-\tan ^{3} \alpha}{1-3 \tan ^{2} \alpha}$, we have $t_{k+1}=\frac{3 t_{k}-t_{k}^{3}}{1-3 t_{k}^{2}}$
Thus, $\frac{t_{k+1}}{t_{k}}=\frac{3-t_{k}^{2}}{1-3 t_{k}^{2}}=\frac{\sqrt{3}+t_{k}}{1-\sqrt{3} t_{k}} \cdot \frac{\sqrt{3}-t_{k}}{1+\sqrt{3} t_{k}}=u_{k} \cdot v_{k}$
Therefore, $\prod_{k=1}^{n} u_{k} v_{k}=\frac{t_{2}}{t_{1}} \cdot \frac{t_{3}}{t_{2}} \cdots \frac{t_{n}}{t_{n-1}} \cdot \frac{t_{n+1}}{t_{n}}=\frac{t_{n+1}}{t_{1}}=\frac{\tan \frac{3^{n} \pi}{3^{n}-1}}{\tan \frac{\pi}{3^{n}-1}}=1$
So, $\prod_{k=1}^{n} u_{k}=\frac{1}{\prod_{k=1}^{n} v_{k}}=\prod_{k-1}^{n} \frac{1}{v_{k}}$
Thus, $\prod_{k=1}^{n} \tan \left[\frac{\pi}{3}\left(1+\frac{3^{k}}{3^{n}-1}\right)\right]=\prod_{k=1}^{n} \cot \left[\frac{\pi}{3}\left(1-\frac{3^{k}}{3^{n}-1}\right)\right]$