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Question

(A) The given differential equation is: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$.Using trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ a

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

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(A) The given differential equation is: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$.Using trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we get:$\frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = 	an^2 \frac{x}{2}$.Since $	an^2 	heta = \sec^2 	heta - 1$,we have:$\frac{dy}{dx} = \sec^2 \frac{x}{2} - 1$.Separating the variables,we get:$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$.Thus,the general solution is $y = 2 	an \frac{x}{2} - x + C$.

Find the general solution of the differential equation: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$

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