Integrating factor of differential equation \([1 /(\cos x)] \cdot(d y / d x)+[1 /(\sin x)] y=1\) is. (A) \(\sec x\) (B) \(\cos \mathrm{x}\) (C) \(\tan \mathrm{x}\) (D) \(\sin \mathrm{x}\)

Given the differential equation \(\frac{1}{\cos x} \frac{dy}{dx} + \frac{1}{\sin x}y = 1\), we can see that it is in the form of a first-order linear differential equation with P(x) = \(\frac{1}{\sin x}\) and Q(x) = 1.

As mentioned earlier, the integrating factor is given by the formula \(e^{\int P(x)dx}\). Let's find \(\int P(x)dx\): \[\int P(x)dx = \int \frac{1}{\sin x} dx\] Let's make a substitution: \(u = \sin x, du = \cos x dx\) \[\int \frac{1}{u} \cdot \frac{du}{\cos x} = \frac{1}{\cos x}\int \frac{1}{u} du\] Now let's integrate: \[ \frac{1}{\cos x}\int \frac{1}{u} du = \frac{ln|u|+c}{\cos x} = \frac{ln|\sin x| + c}{\cos x}\] Now, let's find the integrating factor: \[e^{\int P(x)dx} = e^{\frac{ln|\sin x| + c}{\cos x}}\] But we know that \(e^{\ln z} = z\), so: \[e^{\int P(x)dx} = \frac{e^{ln|\sin x|+c}}{\cos x} = \frac{\sin x}{\cos x}\] Thus, the integrating factor is \(\tan x\), which corresponds to the answer choice (C).