The solution of $\frac{dy}{dx}=\cos \left(x+y\right)+\sin \left(x+y\right)$, is

Put $x+y=v$ and $1+\frac{dy}{dx}=\frac{dv}{dx}$

Therefore, the differential equation reduces to

$\frac{dv}{dx}=\left(1+\cos v\right)+\sin v=2{\cos }^2\frac{v}{2}+2\sin \frac{v}{2}\cos \frac{v}{2}$

$=2{\cos }^2\frac{v}{2}\left(1+\frac{\tan v}{2}\right)$

$\Rightarrow \int \frac{{\sec }^2\frac{V}{2}}{2\left[1+\tan \left(\frac{v}{2}\right)\right]}dv=\int dx$

$\therefore \log \left[1+\tan \left(\frac{x+y}{2}\right)\right]=x+C$