1. 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
1. A Prompt: Left side $=\sin A \cos A+\sin A \cos B+\cos A \sin B+\sin B \cos B$
$$
\begin{array}{l}
=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B) \\
=\sin (A+B) \cos (A-B)+\sin (A+B)
\end{array}
$$

Right side $=2 \sin \left[180^{\circ}-(A+B)\right]=2 \sin (A+B)$
Thus, $\sin (A+B) \cos (A-B)+\sin (A+B)=2 \sin (A+B)$
which simplifies to $\sin (A+B)[\cos (A-B)-1]=0$
Since $\sin (A+B)>0$
it follows that $\cos (A-B)=1$, so $A=B$
Taking $A=B=30^{\circ}, C=120^{\circ}$ 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.