12.361. A regular $n$-sided pyramid is circumscribed around a sphere of radius $R$, with the lateral face forming an angle $\alpha$ with the base plane. Find the lateral surface area of the pyramid.
Solution.

The lateral surface area of a regular pyramid $S_{\bar{o}}=\frac{S}{\cos \alpha}$, where $S$ is the area of the base, $\alpha$ is the angle between the lateral face and the base plane.

$B O_{1}$ is the height of the given regular pyramid $B A_{1} A_{2} \ldots A_{n^{\prime}} B E$ is the apothem of its lateral face $A_{1} B A_{2}, \angle B E O_{1}=\alpha$ (Fig.12.221).

The center $O$ of the inscribed sphere lies on the height $B O_{1}, O O_{1}=R$, $\angle O E O_{1}=\frac{1}{2} \angle B E O_{1}=\frac{\alpha}{2}$.

In $\triangle O O_{1} E\left(\angle O O_{1} E=90^{\circ}\right): O_{1} E=R \operatorname{ctg} \frac{\alpha}{2}$.

In $\triangle A_{1} E O_{1}\left(\angle A_{1} E O_{1}=90^{\circ}\right): A_{1} E=O_{1} E \operatorname{tg} \angle A_{1} O_{1} E=R \operatorname{ctg} \frac{\alpha}{2} \operatorname{tg} \frac{\pi}{n}$.

Then

$$
\begin{aligned}
& S=n S_{\triangle A_{1} O_{1} A_{2}}=n \cdot A_{1} E \cdot O_{1} E=n R^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \operatorname{tg} \frac{\pi}{n} \\
& S_{\sigma}=\frac{n R^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \operatorname{tg} \frac{\pi}{n}}{\cos \alpha}
\end{aligned}
$$

Answer: $\frac{n R^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \operatorname{tg} \frac{\pi}{n}}{\cos \alpha}$.