3.90. In a truncated cone, the diagonals of the axial section are mutually perpendicular and the length of each is a. The angle between the generatrix and the base plane is $\alpha$. Find the total surface area of the truncated cone.
3.90. Let's depict the axial section of a truncated cone. According to the problem, \( A B_{1} \perp A_{1} B \), \( O O_{1} \) is the axis of the truncated cone, \( B B_{1} \) is the generatrix of the truncated cone, \( A B_{1} = a \). \(\angle B_{1} B O = \alpha\) (Fig. 3.90). Let \( O B = R_{1} \), \( O_{1} B_{1} = R_{2} \); then \( M O = O B = R_{1}, M O_{1} = O_{1} B_{1} = R_{2}, O O_{1} = R_{1} + R_{2} \). Draw \( B_{1} C \parallel O O_{1} \); in this case, \( B_{1} C = O_{1} O = R_{1} + R_{2} \). In \(\triangle B_{1} C A\), we have \(\angle B_{1} A C = 45^{\circ}\); therefore, \( B_{1} C = A C = R_{1} + R_{2} = \frac{a \sqrt{2}}{2} \). Further, from \(\triangle B_{1} C B\), we find

\[
B_{1} B = \frac{B_{1} C}{\sin \alpha} = \frac{a \sqrt{2}}{2 \sin \alpha}, \quad B C = R_{1} - R_{2} = B_{1} C \operatorname{ctg} \alpha = \frac{a \sqrt{2}}{2} \operatorname{ctg} \alpha
\]

Thus, \( S_{\text{lat}} = \pi (R_{1} + R_{2}) \cdot B_{1} B = \frac{\pi a^{2}}{2 \sin \alpha} \). Solving the system of equations

\[
R_{1} + R_{2} = \frac{a \sqrt{2}}{2}, \quad R_{1} - R_{2} = \frac{a \sqrt{2}}{2 \operatorname{ctg} \alpha}
\]

we find

\[
R_{1} = \frac{a \sqrt{2}}{4} (1 + \operatorname{ctg} \alpha) = \frac{a \sin \left(\frac{\pi}{4} + \alpha\right)}{2 \sin \alpha}, \quad R_{2} = \frac{a \sqrt{2}}{4} (1 - \operatorname{ctg} \alpha) = \frac{a \sin \left(\frac{\pi}{4} - \alpha\right)}{2 \sin \alpha}.
\]

Thus,

\[
\begin{aligned}
& S_{\text{full}} = \frac{\pi a^{2}}{2 \sin \alpha} + \frac{\pi a^{2} \sin^{2} \left(\frac{\pi}{4} + \alpha\right)}{4 \sin^{2} \alpha} + \frac{\pi a^{2} \sin^{2} \left(\frac{\pi}{4} - \alpha\right)}{4 \sin^{2} \alpha} = \\
& \frac{\pi a^{2}}{4 \sin^{2} \alpha} \left(2 \sin \alpha + \frac{1 - \cos \left(\frac{\pi}{2} + 2 \alpha\right) + 1 - \cos \left(\frac{\pi}{2} - 2 \alpha\right)}{2}\right) = \\
& \frac{\pi a^{2}}{4 \sin^{2} \alpha} (2 \sin \alpha + 1) = \frac{\pi a^{2}}{\sin^{2} \alpha} \sin \left(\frac{\alpha}{2} + \frac{\pi}{12}\right) \cos \left(\frac{\alpha}{2} - \frac{\pi}{12}\right)
\end{aligned}
\]