Task B-4.5. If $\cos 4 \alpha=\frac{3}{\sqrt{5 \sqrt{5 \sqrt{5 \sqrt{5 \cdots}}}}}$, what is $(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}$?
Solution. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$ is a geometric series with $a_{1}=\frac{1}{2}, q=\frac{1}{2}<1$ so its sum is

$$
S=\frac{1}{2} \cdot \frac{1}{1-\frac{1}{2}}=1
$$

Then we have

$$
\begin{gathered}
\cos 4 \alpha=\frac{3}{\sqrt{5 \sqrt{5 \sqrt{5 \sqrt{5 \cdots}}}}}=\frac{3}{5^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots}}=\frac{3}{5} \\
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}=\left(\frac{1}{\sin \alpha \cos \alpha}\right)^{2}=\frac{4}{\sin ^{2} 2 \alpha}=\frac{8}{1-\cos 4 \alpha}=\frac{8}{1-\frac{3}{5}}=20
\end{gathered}
$$