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(B) Given,$\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$. Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}

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$x + \operatorname{cosec}(x+y) = c$

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(B) Given,$\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$. Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,so $\frac{dy}{dx} = \frac{dz}{dx} - 1$. Substituting this into the differential equation: $\frac{dz}{dx} - 1 = \sin z 	an z - 1$ $\frac{dz}{dx} = \sin z 	an z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$. Rearranging the variables: $\int \frac{\cos z}{\sin^2 z} dz = \int dx$. Let $\sin z = t$,then $\cos z dz = dt$. $\int \frac{1}{t^2} dt = x + c$ $- \frac{1}{t} = x + c$ $- \operatorname{cosec} z = x + c$ Substituting $z = x+y$ back: $- \operatorname{cosec}(x+y) = x + c$ $x + \operatorname{cosec}(x+y) = c$ (where $c$ is an arbitrary constant).

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

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