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(C) Given the differential equation: $(1+y^2) e^{	an x} dx + \cos^2 x(1+e^{2 	an x}) dy = 0$.Rearranging the terms to separate variables:$\frac{e^{\

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$\frac{1}{e}$

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(C) Given the differential equation: $(1+y^2) e^{	an x} dx + \cos^2 x(1+e^{2 	an x}) dy = 0$.Rearranging the terms to separate variables:$\frac{e^{	an x}}{\cos^2 x(1+e^{2 	an x})} dx + \frac{dy}{1+y^2} = 0$.Since $\sec^2 x = \frac{1}{\cos^2 x}$,we have:$\frac{\sec^2 x e^{	an x}}{1+(e^{	an x})^2} dx + \frac{dy}{1+y^2} = 0$.Integrating both sides:$\int \frac{\sec^2 x e^{	an x}}{1+(e^{	an x})^2} dx + \int \frac{dy}{1+y^2} = C$.Let $u = e^{	an x}$,then $du = e^{	an x} \sec^2 x dx$. The integral becomes:$\int \frac{du}{1+u^2} + \int \frac{dy}{1+y^2} = C$.$	an^{-1}(e^{	an x}) + 	an^{-1}(y) = C$.Given $y(0) = 1$,substitute $x=0$ and $y=1$:$	an^{-1}(e^{	an 0}) + 	an^{-1}(1) = C \implies 	an^{-1}(1) + 	an^{-1}(1) = C \implies \frac{\pi}{4} + \frac{\pi}{4} = C \implies C = \frac{\pi}{2}$.So,$	an^{-1}(e^{	an x}) + 	an^{-1}(y) = \frac{\pi}{2}$.We know $	an^{-1}(A) + \cot^{-1}(A) = \frac{\pi}{2}$,so $	an^{-1}(y) = \cot^{-1}(e^{	an x}) = 	an^{-1}(\frac{1}{e^{	an x}})$.Thus,$y = \frac{1}{e^{	an x}}$.For $x = \frac{\pi}{4}$,$y(\frac{\pi}{4}) = \frac{1}{e^{	an(\pi/4)}} = \frac{1}{e^1} = \frac{1}{e}$.

Let $y=y(x)$ be the solution of the differential equation $(1+y^2) e^{\tan x} dx + \cos^2 x(1+e^{2 \tan x}) dy = 0$,with $y(0)=1$. Then $y(\frac{\pi}{4})$ is equal to:

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