Assertion (A) : Integrating factor of $1+\mathrm{y}(\mathrm{x} \tan \mathrm{y}-\sec \mathrm{y}) \frac{\mathrm{dy}}{\mathrm{dx}}=0$ is $\sec \mathrm{y}$

Reason (R): Integrating factor of $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{Q}\left(\mathrm{x}\right)$ is ${\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}$

Then the Statement among the following is

$\begin{array}{rcl}1+\left(\mathrm{x} \tan \mathrm{y}-\sec \mathrm{y}\right)\frac{\mathrm{dy}}{\mathrm{dx}}& =& 0\\ & \Rightarrow & \left(\mathrm{x} \tan \mathrm{y}-\sec \mathrm{y}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=-1\\ & \Rightarrow & \mathrm{x} \tan \mathrm{y}-\sec \mathrm{y}=\frac{-\mathrm{dy}}{\mathrm{dx}}\\ & \Rightarrow & \frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{x} \tan \mathrm{y}=\sec \mathrm{y}\\ \mathrm{P}& =& \tan \mathrm{y}, \mathrm{Q}=\sec \mathrm{y}\\ \mathrm{I}.\mathrm{F}& =& {\mathrm{e}}^{\int \tan \mathrm{y} \mathrm{dy}}={\mathrm{e}}^{\int \log \mathrm{P}\left(\mathrm{x}\right)}\\ \mathrm{I}.\mathrm{F}& =& \sec \mathrm{y}\end{array}$

$\therefore$ Assertion is true. So both A & R are ture and is r is correct explanation A.