The general solution of the differential equation \(\cos (x+y) \frac{d y}{d x}=1\) is

\(
\cos (x+y) \frac{d y}{d x}=1
\)
Put \(x+y=V \Rightarrow 1+\frac{d y}{d x}=\frac{d V}{d x}\)
\( \therefore \cos \mathrm{V}\left(\frac{\mathrm{dV}}{\mathrm{dx}}-1\right)=1 \Rightarrow \cos \mathrm{V}\left(\frac{\mathrm{dV}}{\mathrm{dx}}\right)=1+\cos \mathrm{V} \)
\( \therefore \int \frac{\cos \mathrm{V}}{1+\cos \mathrm{V}} \mathrm{dV}=\int \mathrm{dx} \)
\( \therefore \int\left[\frac{1+\cos \mathrm{V}}{1+\cos \mathrm{V}}-\frac{1}{1+\cos \mathrm{V}}\right] \mathrm{dV} \)
\( =\int \mathrm{dx} \Rightarrow \int \mathrm{dV}-\frac{1}{2} \int \sec ^2 \frac{\mathrm{V}}{2} \mathrm{dV}=\int \mathrm{dx}\)
\( \therefore \mathrm{V}-\frac{1}{2} \frac{\tan \left(\frac{\mathrm{V}}{2}\right)}{\left(\frac{1}{2}\right)}=\mathrm{x}+\mathrm{c} \Rightarrow \mathrm{V}-\tan \left(\frac{\mathrm{V}}{2}\right)=\mathrm{x}+\mathrm{c} \)
\( \Rightarrow \mathrm{x}+\mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\mathrm{x}+\mathrm{c} \)
\( \mathrm{y}=\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)+\mathrm{c}\)