$$
\begin{array}{l}
\quad \text { 4. In } \triangle A B C \text {, if }(\sin A+\sin B)(\cos A+\cos B) \\
=2 \sin C \text {, then } \triangle A B C(\quad .
\end{array}
$$
4. In $\triangle A B C$, if $(\sin A+\sin B)(\cos A+\cos B)$ $=2 \sin C$, then $\triangle A B C(\quad)$.
(A) is an isosceles triangle but not necessarily a right triangle
(B) is a right triangle but not necessarily an isosceles triangle
(C) is neither an isosceles triangle nor a right triangle
(D) is both an isosceles triangle and a right triangle
4. (A).
$$
\begin{array}{l}
\text { Left side }=\sin A \cdot \cos A+\sin A \cdot \cos B+\cos A \cdot \sin B+ \\
\sin B \cdot \cos B=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B) \\
=\sin (A+B) \cos (A-B)+\sin (A+B). \\
\text { Right side }=2 \sin \left[180^{\circ}-(A+B)\right]=2 \sin (A+B). \\
\quad \text { Therefore, } \sin (A+B) \cos (A-B)+\sin (A+B) \\
=2 \sin (A+B),
\end{array}
$$

Left side $=\sin A \cdot \cos A+\sin A \cdot \cos B+\cos A \cdot \sin B+$ $\sin B \cdot \cos B=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B)$ $=\sin (A+B) \cos (A-B)+\sin (A+B)$.
Right side $=2 \sin \left[180^{\circ}-(A+B)\right]=2 \sin (A+B)$.
Therefore, $\sin (A+B) \cos (A-B)+\sin (A+B)$ $=2 \sin (A+B)$,
which means $\sin (A+B)[\cos (A-B)-1]=0$.
Since $\sin (A+B)>0$, we have $\cos (A-B)=1$, hence $A=B$.
Taking $\angle A=\angle B=30^{\circ}, \angle C=120^{\circ}$ and substituting into the condition, we find that it satisfies the condition.

Therefore, $\triangle A B C$ is an isosceles triangle, but not necessarily a right triangle.