3.53. A sphere is inscribed in a cone. The radius of the circle where the cone and the sphere touch is $r$. Find the volume of the cone if the angle between the height and the slant height of the cone is $\alpha$.
3.53. Let's depict the axial section of the figure. In the section, we get $\triangle SDE$, where $SD$ and $SE$ are the generators of the cone, $ED$ is the diameter of its base, $SB$ is the height of the cone, $O$ is the center of the largest circle inscribed in the cone by the sphere, $C$ is the point of tangency of the circle with $SD$, and $A$ is the center of the circle where the sphere touches the lateral surface of the cone (Fig. 3.55). By the condition, $AC = r$, $\angle BSD = \alpha$. Connecting points $O$ and $C$, we have $OC \perp SD$, $AC \perp SB$; therefore, $\angle OCA = \angle BSD = \alpha$. From $\triangle OAC$ we find $OC = \frac{r}{\cos \alpha}$, and from $\triangle OCS$ we get $SO = \frac{OC}{\sin \alpha} = \frac{r}{\sin \alpha \cos \alpha}$. Thus,

\[
\begin{aligned}
SB = & SO + OB = SO + OC = \frac{r}{\sin \alpha \cos \alpha} + \frac{r}{\cos \alpha} = \\
& = \frac{r}{\sin \alpha \cos \alpha}(1 + \sin \alpha) = \frac{2r(1 + \sin \alpha)}{\sin 2\alpha}
\end{aligned}
\]

Next, from $\triangle SBD$ we obtain

\[
BD = SB \operatorname{tg} \alpha = \frac{2r(1 + \sin \alpha) \operatorname{tg} \alpha}{\sin 2\alpha} = \frac{r(1 + \sin \alpha)}{\cos^2 \alpha}
\]

Thus,

\[
\begin{gathered}
V_{\text{dom}} = \frac{1}{3} \pi BD^2 \cdot SB = \frac{1}{3} \pi \frac{r^2(1 + \sin \alpha)^2}{\cos^4 \alpha} \cdot \frac{r(1 + \sin \alpha)}{\sin \alpha \cos \alpha} = \\
= \frac{\pi r^3 (1 + \sin \alpha)^3}{3 \sin \alpha \cos^5 \alpha} = \frac{\pi r^3}{3 \sin \alpha \cos^2 \alpha} \left( \frac{1 - \cos \left( \frac{\pi}{2} + \alpha \right)}{\sin \left( \frac{\pi}{2} + \alpha \right)} \right)^3 = \frac{\pi r^3 \operatorname{tg}^3 \left( \frac{\pi}{4} + \frac{\alpha}{2} \right)}{3 \sin \alpha \cos^2 \alpha}
\end{gathered}
\]