The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})\tan (\mathrm{x}+\mathrm{y})-1:$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})\mathrm{Tan}(\mathrm{x}+\mathrm{y})-1$$\to \left(1\right)$

Put x + y = t 

difff w.r.to x on both sides

1+$\frac{dy}{dx}=\frac{dt}{dx}$

$$\begin{gathered} \left(1\right)\Rightarrow \\ \frac{dt}{dx}=\sin t\tan t \end{gathered}$$

By separating  the variables and integrating we get

$$\begin{gathered} \int \cos ectcott dt=\int dx \\ -\cos ect=x+c \\ \Rightarrow x+\cos ec(x+y)=c \end{gathered}$$