$\frac{dy}{dx}=y\tan x-2\sin x$ then 

The given differential equation is

$\begin{array}{r}\frac{dy}{dx}=y\tan x-2\sin x\\ \Rightarrow \frac{dy}{dx}+(-\tan x)y=-2\sin x\end{array}$                  (i)

which is a linear differential equation

Thus, $\mathrm{IF}=e^{-\int \tan xdx}=e^{-\log (\sec x)}=\cos x$

Multiplying both sides of Eq. (i) by IF and integrating we get

$\begin{array}{l}y\cdot \left(\mathrm{IF}\right)=\int Q\cdot \left(\mathrm{IF}\right)dx+c\\ \Rightarrow y\cdot \cos x=\int (-2\sin x\cdot \cos x)dx+c\\ =-\int \sin 2xdx+c\\ =\frac{\cos 2x}{2}+c\\ \end{array}$

which is the required solution.