$94 \triangle A B C$'s three interior angles satisfy
$$
\sin A \sin B+\sin A \cos B+\cos A \sin B+\cos A \cos B=2
$$

Then the shape of $\triangle A B C$ is
A. Isosceles triangle
B. Right triangle
C. Isosceles right triangle
D. Right or isosceles triangle
94 C. The original expression is transformed into $\sin (A+B)+\cos (A-B)=2$. Since
$$
\sin (A+B) \leqslant 1, \cos (A-B) \leqslant 1,
$$

the above equation holds only when $\sin (A+B)=\cos (A-B)=1$, which means the triangle is an isosceles right triangle.