Solution of $\frac{d y}{d x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is :

$\log \tan \frac{y}{4}=c-2 \sin \frac{x}{2}$

$\frac{d y}{d x}=-2 \cos \frac{x}{2} \sin \frac{y}{2}$

$-\int 2 \cos \frac{x}{2} d x=\int ~cosec \frac{y}{2} d y \Rightarrow c-2 \sin \frac{x}{2}=\log \tan \frac{y}{4}$

Hence (1) is the correct answer.