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(A) Given the differential equation: $(1 + \cos x)dy = (1 - \cos x)dx$Rearranging the terms to separate the variables,we get:$dy = \frac{1 - \cos x}{1

Vedclass · Accepted Answer

$y = 2	an \frac{x}{2} - x + c$

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(A) Given the differential equation: $(1 + \cos x)dy = (1 - \cos x)dx$Rearranging the terms to separate the variables,we get:$dy = \frac{1 - \cos x}{1 + \cos x} dx$Using the trigonometric identities $1 - \cos x = 2\sin^2 \frac{x}{2}$ and $1 + \cos x = 2\cos^2 \frac{x}{2}$,we have:$dy = \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx = 	an^2 \frac{x}{2} dx$Using the identity $	an^2 	heta = \sec^2 	heta - 1$,we get:$dy = (\sec^2 \frac{x}{2} - 1) dx$Integrating both sides:$\int dy = \int (\sec^2 \frac{x}{2} - 1) dx$$y = 2	an \frac{x}{2} - x + c$

The solution of the differential equation $(1 + \cos x)dy = (1 - \cos x)dx$ is

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