If cos x frac{dy}{dx} = y sin x - 1, where x | vedclass

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(A) Given the differential equation: $\cos x \frac{dy}{dx} = y \sin x - 1$.Dividing by $\cos x$, we get: $\frac{dy}{dx} = y 	an x - \sec x$.This is a

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$(1-x) \sec x$

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(A) Given the differential equation: $\cos x \frac{dy}{dx} = y \sin x - 1$.Dividing by $\cos x$, we get: $\frac{dy}{dx} = y 	an x - \sec x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -	an x$ and $Q(x) = -\sec x$.The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int -	an x dx} = e^{\ln(\cos x)} = \cos x$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.Substituting the values: $y \cos x = \int (-\sec x) \cdot \cos x dx + C$.$y \cos x = \int (-1) dx + C$.$y \cos x = -x + C$.Given $f(0) = 1$, we substitute $x = 0$ and $y = 1$: $1 \cdot \cos(0) = -0 + C \implies 1 = C$.Thus, $y \cos x = -x + 1$.$y = \frac{1-x}{\cos x} = (1-x) \sec x$.Therefore, $f(x) = (1-x) \sec x$.

If $\cos x \frac{dy}{dx} = y \sin x - 1$, where $x \neq (2n+1) \frac{\pi}{2}, n \in Z$, is the differential equation corresponding to the curve $y = f(x)$ and $f(0) = 1$, then $f(x) =$

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