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(A) Given differential equation: $e^x 	an y \, dx + (1+e^x) \sec^2 y \, dy = 0$Rearranging the terms: $e^x 	an y \, dx = -(1+e^x) \sec^2 y \, dy$Sep

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$(1+e^x) 	an y = 2$

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(A) Given differential equation: $e^x 	an y \, dx + (1+e^x) \sec^2 y \, dy = 0$Rearranging the terms: $e^x 	an y \, dx = -(1+e^x) \sec^2 y \, dy$Separating the variables: $\frac{e^x}{1+e^x} \, dx = -\frac{\sec^2 y}{	an y} \, dy$Integrating both sides: $\int \frac{e^x}{1+e^x} \, dx = -\int \frac{\sec^2 y}{	an y} \, dy$Using the formula $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$,we get:$\ln(1+e^x) = -\ln(	an y) + \ln C$$\ln(1+e^x) + \ln(	an y) = \ln C$$\ln[(1+e^x) 	an y] = \ln C$$(1+e^x) 	an y = C$Since the curve passes through the point $\left(0, \frac{\pi}{4}ight)$,substitute $x=0$ and $y=\frac{\pi}{4}$:$(1+e^0) 	an(\frac{\pi}{4}) = C$$(1+1)(1) = C \implies C = 2$Thus,the equation of the curve is $(1+e^x) 	an y = 2$.

The equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ and satisfying the differential equation $\left(e^x \tan y\right) dx + \left((1+e^x) \sec^2 y\right) dy = 0$ is given by

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