Integrating factor of the differential equation dy/dx + y tan x – sec x = 0 is

The correct option is sec x.

To find the integrating factor of the given differential equation, we first write it in standard form.

The given first-order differential equation is:
$$\frac{dy}{dx}+y\tan x-\sec x=0$$

Rearranging the terms, we get:
$$\frac{dy}{dx}+(\tan x)y=\sec x$$

This is a linear differential equation of the form:
$$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$$

By comparison, we identify the coefficient functions:
$$P\left(x\right)=\tan x$$

The formula for the integrating factor (I.F.) is:
$$\text{I.F.}=e^{\int P\left(x\right)dx}$$

Substituting $P\left(x\right)=\tan x$ into the formula:
$$\text{I.F.}=e^{\int \tan xdx}$$

Since the integration of the tangent function is given by:
$$\int \tan xdx=\ln |\sec x|$$

We can substitute this back into the exponent:
$$\text{I.F.}=e^{\ln |\sec x|}$$

Using the identity $e^{\ln u}=u$, the expression simplifies to:
$$\text{I.F.}=\sec x$$

Thus, the integrating factor is sec x.