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The solution of the differential equation $\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1$ is
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The correct answer is:
$x+\operatorname{cosec}(x+y)=c$
Given, $\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1$
$\begin{aligned} & \text { Put } x+y=z \Rightarrow \quad 1+\frac{d y}{d x}=\frac{d z}{d x} \\ & \therefore \quad \frac{d z}{d x}-1=\sin z \tan z-1 \\ & \Rightarrow \quad \int \frac{\cos z}{\sin ^2 z} d z=\int d x\end{aligned}$
$\begin{aligned} & \text { Put } \quad \sin z=t \\ & \Rightarrow \quad \cos z d z=d t \\ & \therefore \quad \int \frac{1}{t^2} d t=x-c \Rightarrow-\frac{1}{t}=x-c \\ & \Rightarrow \quad-\operatorname{cosec} z=x-c \\ & \Rightarrow \quad x+\operatorname{cosec}(x+y)=c \\ & \end{aligned}$
$\begin{aligned} & \text { Put } x+y=z \Rightarrow \quad 1+\frac{d y}{d x}=\frac{d z}{d x} \\ & \therefore \quad \frac{d z}{d x}-1=\sin z \tan z-1 \\ & \Rightarrow \quad \int \frac{\cos z}{\sin ^2 z} d z=\int d x\end{aligned}$
$\begin{aligned} & \text { Put } \quad \sin z=t \\ & \Rightarrow \quad \cos z d z=d t \\ & \therefore \quad \int \frac{1}{t^2} d t=x-c \Rightarrow-\frac{1}{t}=x-c \\ & \Rightarrow \quad-\operatorname{cosec} z=x-c \\ & \Rightarrow \quad x+\operatorname{cosec}(x+y)=c \\ & \end{aligned}$
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