Solution of the differential equation  $\frac{dy}{dx}=y\tan x-y^2\sec x \text{ is }$

$\frac{dy}{dx}=y\tan x-y^2\sec x$

$\Rightarrow -\frac{1}{y^2}\frac{dy}{dx}+\frac{\tan x}{y}=\sec x\Rightarrow \frac{dz}{dx}+\tan x\cdot z=\sec x,z=\frac{1}{y}$

$\text{ I.F. }=e^{\int \tan x dx}=e^{\ln \sec x}=\sec x$

$\therefore \text{ Solution is, }z\sec x=\int {\sec }^{2}xdx+c$

$\frac{\sec x}{y}=\tan x+c\Rightarrow \sec x=y(\tan x+c)$