2.3.3 * In $\triangle A B C$, if $(\sin A+\sin B)(\cos A+\cos B)=2 \sin C$, then ( ).
(A) $\triangle A B C$ is an isosceles triangle, but not necessarily a right triangle
(B) $\triangle A B C$ is a right triangle, but not necessarily an isosceles triangle
(C) $\triangle A B C$ is neither an isosceles triangle nor a right triangle
(D) $\triangle A B C$ is both an isosceles triangle and a right triangle
Given the left side of the equation $=\sin A \cdot \cos A + \sin A \cdot \cos B + \sin B \cdot \cos A + \sin B \cdot \cos B = \frac{1}{2}(\sin 2A + \sin 2B) + \sin (A+B) = \sin (A+B) \cdot \cos (A-B) + \sin (A+B)$, and the right side $=2 \sin (A+B)$. Therefore, the given equation can be transformed into $\sin (A+B)[\cos (A-B) - 1] = 0$.
Since $\sin (A+B) > 0$, it follows that $\cos (A-B) = 1$. Hence, $\angle A = \angle B$.
On the other hand, $\angle A = \angle B = 30^\circ, \angle C = 120^\circ$ also satisfies the given conditions. Therefore, $\triangle ABC$ is an isosceles triangle, but not necessarily a right triangle. Hence, the answer is A.