4. If the cosines of the three interior angles of $\triangle A_{1} B_{1} C_{1}$ are the sines of the three interior angles of $\triangle A_{2} B_{2} C_{2}$, then, ( . . .
(A) $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ are both acute triangles
(B) $\triangle A_{1} B_{1} C_{1}$ is an acute triangle, $\triangle A_{2} B_{2} C_{2}$ is an obtuse triangle
(C) $\triangle A_{1} B_{1} C_{1}$ is an obtuse triangle, $\triangle A_{2} B_{2} C_{2}$ is an acute triangle
(D) $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ are both obtuse triangles
4.B.

Since the interior angles of the two triangles cannot be right angles, and the cosine values of the interior angles of $\triangle A_{1} B_{1} C_{1}$ are all greater than zero, $\triangle A_{1} B_{1} C_{1}$ is an acute triangle.
If $\triangle A_{2} B_{2} C_{2}$ is an acute triangle, without loss of generality, we have
$$
\begin{array}{l}
\cos A_{1}=\sin A_{2}=\cos \left(\frac{\pi}{2}-A_{1}\right), \\
\cos B_{1}=\sin B_{2}=\cos \left(\frac{\pi}{2}-B_{1}\right), \\
\cos C_{1}=\sin C_{2}=\cos \left(\frac{\pi}{2}-C_{1}\right) . \\
\text { Then } \angle A_{1}=\frac{\pi}{2}-\angle A_{2}, \angle B_{1}=\frac{\pi}{2}-\angle B_{2}, \\
\angle C_{1}=\frac{\pi}{2}-\angle C_{2},
\end{array}
$$

That is
$$
\begin{array}{l}
\angle A_{1}+\angle B_{1}+\angle C_{1} \\
=\frac{3 \pi}{2}-\left(\angle A_{1}+\angle B_{1}+\angle C_{1}\right)
\end{array}
$$

Contradiction.