The general solution of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x\) is

\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x \\
& \Rightarrow \frac{1}{y^2} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}+\tan x\left(-\frac{1}{y}\right)=-\sec x \ldots (i)\\
& \text { Put } \mathrm{v}=-\frac{1}{y} \Rightarrow \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{1}{y^2} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}
\end{aligned}\)
\(\therefore \quad \frac{\mathrm{dv}}{\mathrm{d} x}+(\tan x) \mathrm{v}=-\sec x \quad \ldots[\) From (i) \(]\)
\(\therefore \quad\) I.F. \(=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log (\sec x)}=\sec x\)
\(\therefore \quad\) solution of the given equation is
\(\begin{aligned} & \text { v. } \sec x=\int-\sec x \cdot \sec x \mathrm{~d} x+\mathrm{c}_1 \\ & \Rightarrow \mathrm{v} \sec x=-\tan x+\mathrm{c}_1 \\ & \Rightarrow-\frac{1}{y} \sec x=-\tan x+\mathrm{c}_1 \\ & \Rightarrow \sec x=y(\tan x+\mathrm{c}), \text { where } \mathrm{c}=-\mathrm{c}_1\end{aligned}\)