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(B) Given the slope of the tangent is $\frac{dy}{dx} = \cos(x + y) + \sin(x + y)$.Let $u = x + y$,then $\frac{du}{dx} = 1 + \frac{dy}{dx}$,so $\frac{d

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$y = 2 	an^{-1}(e^x - 1) - x$

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(B) Given the slope of the tangent is $\frac{dy}{dx} = \cos(x + y) + \sin(x + y)$.Let $u = x + y$,then $\frac{du}{dx} = 1 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{du}{dx} - 1$.Substituting this into the equation: $\frac{du}{dx} - 1 = \cos u + \sin u$.$\frac{du}{dx} = 1 + \cos u + \sin u = 2 \cos^2(\frac{u}{2}) + 2 \sin(\frac{u}{2}) \cos(\frac{u}{2}) = 2 \cos^2(\frac{u}{2}) [1 + 	an(\frac{u}{2})]$.Separating variables: $\int \frac{\sec^2(\frac{u}{2})}{2(1 + 	an(\frac{u}{2}))} du = \int dx$.Let $t = 	an(\frac{u}{2})$,then $dt = \frac{1}{2} \sec^2(\frac{u}{2}) du$.The integral becomes $\int \frac{dt}{1 + t} = \int dx$,which gives $\ln(1 + t) = x + C$.Since the curve passes through the origin $(0, 0)$,$u = x + y = 0$,so $t = 	an(0) = 0$.Substituting $x = 0, t = 0$ into $\ln(1 + t) = x + C$ gives $C = 0$.Thus,$\ln(1 + t) = x$,which implies $1 + t = e^x$,or $t = e^x - 1$.Substituting back $t = 	an(\frac{x + y}{2})$,we get $	an(\frac{x + y}{2}) = e^x - 1$.Therefore,$\frac{x + y}{2} = 	an^{-1}(e^x - 1)$,which simplifies to $y = 2 	an^{-1}(e^x - 1) - x$.

The equation of a curve passing through the origin,if the slope of the tangent drawn at any of its points $(x, y)$ is $\cos (x + y) + \sin (x + y)$,is

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