64. In a sphere of radius $R$, a diameter $A B$ is drawn. Two lines are tangent to the sphere at points $A$ and $B$ and form an angle $\alpha\left(\alpha<90^{\circ}\right)$ between them. Points $C$ and $D$ are taken on these lines such that $C D$ is also tangent to the sphere and the angle between $A B$ and $C D$ is $\varphi\left(\varphi<90^{\circ}\right)$. Find the volume of the tetrahedron $A B C D$.
64. Let $|A C|=x,|B D|=y$ (where $A C$ and $B D$ are tangent to the sphere). $D_{1}$ is the projection of $D$ onto the plane passing through $A C$ parallel to $B D$. We have

$$
|C D|=x+y=\frac{2 R}{\cos \varphi}, \quad\left|C D_{1}\right|=2 R \operatorname{tg} \varphi
$$

In $\triangle C A D_{1}$, the angle $\widehat{C A D}_{1}$ is equal to $\alpha$ or $180^{\circ}-\alpha$. Accordingly, $x$ and $y$ must satisfy one of the two systems of equations

$$
\left\{\begin{array}{l}
x+y=\frac{2 R}{\cos \varphi} \\
x^{2}+y^{2}-2 x y \cos \alpha=4 R^{2} \operatorname{tg}^{2} \varphi
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x+y=\frac{2 R}{\cos \varphi} \\
x^{2}+y^{2}+2 x y \cos \alpha=4 R^{2} \operatorname{tg}^{2} \varphi
\end{array}\right.
$$

For system (1), we obtain: $x+y=\frac{2 R}{\cos \varphi}, \quad x y=\frac{R^{2}}{\cos ^{2} \frac{\alpha}{2}}$; for system (2) $-x+y=\frac{2 R}{\cos \varphi}, \quad x y=\frac{R^{2}}{\sin ^{2} \frac{\alpha}{2}}$. Considering the inequality $(x+y)^{2} \geqslant 4 x y$, we get that system (1) has a solution when $\varphi \geqslant \frac{\alpha}{2}$, and system (2) when $\varphi \geqslant \frac{\pi}{2}-\frac{\alpha}{2}$. Since the volume of the tetrahedron $A B C D$ is equal to $\frac{1^{2}}{3} x y R \sin \alpha$, we get the answer: if $\frac{\alpha}{2} \leqslant \varphi<\frac{\pi}{2}-\frac{\alpha}{2}$, the volume of the tetrahedron is $\frac{2}{3} R^{3} \operatorname{tg} \frac{\alpha}{2}$; if $\frac{\pi}{2}-\frac{\alpha}{2} \leqslant \varphi<\frac{\pi}{2}$, there are two possible values for the volume $\frac{2}{3} R^{3} \operatorname{tg} \frac{\alpha}{2}$ and $\frac{2}{3} R^{3} \operatorname{ctg} \frac{\alpha}{2}$.