Given a right circular cone with a semi-aperture angle $\alpha$. What fraction of the cone's volume is the volume of the sphere inscribed in the cone?
Let the apex of the cone be $A$, the foot of its height be $C$, the center of the sphere be $O$, the radius of the sphere be $\varrho$, the radius of the base circle be $r$, and let $A O=x$.

The volume of the cone: $V_{k}=\frac{r^{2} \pi m}{3}$.

The volume of the sphere: $V_{g}=\frac{4 \varrho^{3} \pi}{3}$.

Thus, $\frac{V_{g}}{V_{k}}=\frac{4 \varrho^{3}}{r^{2} m}$. From the right triangles $A B C$ and $A O D$, we have $r=(x+\varrho) \operatorname{tg} \alpha$; substituting $x=\frac{\varrho}{\sin \alpha}$, we get $r=\operatorname{tg} \alpha\left(\frac{\varrho}{\sin \alpha}+\varrho\right)=\frac{\varrho(\sin \alpha+1)}{\cos \alpha}, \quad m=x+\varrho=\frac{\varrho}{\sin \alpha}+\varrho=\frac{\varrho(\sin \alpha+1)}{\sin \alpha}$,

therefore

$$
\frac{V_{g}}{V_{k}}=\frac{4 \varrho^{3}}{\left[\frac{\varrho(\sin \alpha+1)}{\cos \alpha}\right]^{2} \frac{\varrho(\sin \alpha+1)}{\sin \alpha}}=\frac{4 \cos ^{2} \alpha \sin \alpha}{(1+\sin \alpha)^{3}}=\frac{4 \sin \alpha(1-\sin \alpha)}{(1+\sin \alpha)^{2}}
$$

This completes the solution of the problem.

Remark. The result of the problem could also have been obtained using other trigonometric functions. The different formulas can be transformed into each other using the relationships between trigonometric functions.