Solve: $\cos x\, dy=y(\sin x-y)\,dx$

Question

Solve: $\cos x \,dy=y(\sin x-y)\,dx$

I have done the solution the following way. Is my answer correct or wrong?

The answer given in the book:

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

Answer 1

Given $\cos xdy=y(\sin x-y)dx$ $$\frac{dy}{dx}=y\tan x-y^2\sec x$$$$\dfrac{1}{y^2}\frac{dy}{dx}-\frac1y\tan x=-\sec x.....(1)$$ Let $\dfrac1y=t\implies-\dfrac{1}{y^2}\dfrac{dy}{dx}=\dfrac{dt}{dx}$, then equation $(1)$ becomes$$-\frac{dt}{dx}-t\tan x=-\sec x$$$$\frac{dt}{dx}+t\tan x=\sec x$$ Now,$$I.F=e^{\int\tan xdx}=e^{\log(\sec x)}=\sec x$$ $$t\times I.F=\int\sec x\times I.F.dx$$$$t\sec x=\int\sec^2xdx$$$$t\sec x=\tan x+C$$$$\frac1y\sec x=\tan x+C$$$$\sec x=y(\tan x+C)$$