The solution of the differential equation $\cos xdy=y\left(\sin \right(x)-y)dx,$ $0<x<\frac{\pi }{2}$ is
$($Here, $C$ is a constant of integration$)$

The correct option is D $\sec x=(\tan x+C)y$
$\frac{dy}{dx}=\frac{y\left(\sin \right(x)-y)}{\cos x}\Rightarrow \frac{dy}{dx}-\left(\tan x\right)y=-\left(\sec x\right)y^2\Rightarrow y^{-2}\frac{dy}{dx}+\frac{(-\tan x)}{y}=-\sec x\dots \left(1\right)$
which is Bernoulli's differential equation.
Put $\frac{1}{y}=t$
$\Rightarrow \frac{-1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

From equation $\left(1\right),$
$\frac{dt}{dx}+\left(\tan x\right)t=\sec x$
I.F. $=e^{\int \tan xdx}=e^{\ln \sec x}=\sec x$
So, the general solution is
$t\sec x=\int {\sec }^{2}xdx\Rightarrow t\sec x=\tan x+C$
Hence, $\frac{1}{y}\sec x=\tan x+C$
$\Rightarrow \sec x=(\tan x+C)y$