1. Given differential equation

$\cos x\,dy = y(\sin x-y)\,dx, \qquad 0<x<\frac{\pi}{2}$

We need to solve for $y$.

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2. Rewrite in standard form

Divide both sides by $dx$:

$\cos x\,\frac{dy}{dx}=y\sin x-y^2$

Now divide by $\cos x$:

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

So,

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

This is a Bernoulli-type equation.

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3. Use substitution

Let

$v=\frac{1}{y}$

Then

$\frac{dv}{dx}=-\frac{1}{y^2}\frac{dy}{dx}$

Multiply the differential equation

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

by $-\frac{1}{y^2}$:

$-\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\tan x=\sec x$

Using $v=\frac1y$ and $\frac{dv}{dx}=-\frac1{y^2}\frac{dy}{dx}$,

$\frac{dv}{dx}+v\tan x=\sec x$

This is now a linear differential equation.

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4. Solve the linear equation

$\frac{dv}{dx}+v\tan x=\sec x$

The integrating factor is

$IF=e^{\int \tan x\,dx}=e^{-\ln(\cos x)}=\sec x$

Multiply throughout by $\sec x$:

$\sec x\frac{dv}{dx}+v\sec x\tan x=\sec^2 x$

Left side becomes:

$\frac{d}{dx}(v\sec x)=\sec^2 x$

Integrate:

$v\sec x=\tan x+c$

Substitute back $v=\frac1y$:

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

Rearrange:

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

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5. Match with options

This is exactly:

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

So the correct option is D.

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6. Check other options briefly

• A: $y\sec x=\tan x+c$
  Not same as obtained.
• B: $y\tan x=\sec x+c$
  Not same.
• C: $\tan x=(\sec x+c)y$
  Not same.
• D: $\sec x=(\tan x+c)y$
  Correct.

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Final Answer: $\boxed{\text{D}}$