Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0<x<\frac{\pi}{2}$ is

$\sec x=(\tan x+C) y$

We have,

$\cos x d y=y(\sin x-y) d x$

$\Rightarrow \frac{d y}{d x}=y \tan x-y^2 \sec x$

$\Rightarrow \frac{1}{y^2} \frac{d y}{d x}+\tan x\left(-\frac{1}{y}\right)=-\sec x$

$\Rightarrow \frac{d v}{d x}+(\tan x) v=-\sec x$, where $v=-\frac{1}{y}$

This is a linear differential equation with integrating factor $e^{\int \tan x d x}=\sec x$. So, its solution is given by

$v \sec x=-\int \sec ^2 x d x+C$

$\Rightarrow v \sec x=-\tan x-C$

$\Rightarrow -\frac{1}{y} \sec x=-\tan x-C$

$\Rightarrow \sec x=y(\tan x+C)$