Solution of differential equation $dy- \sin x \sin ydx=0$ is :

Given $dy- \sin x \sin y dx = 0$

$\Rightarrow dy=\sin x\sin ydx$

   Dividing the L.H.S. and R.H.S. by sin y We get:

$\mathrm{cosec}ydy=\sin xdx$

Integrating on both sides we get: 

$\Rightarrow \log \tan y/2=c-\cos x$

Note $:\log x=y\Rightarrow x=e^y$

So, $\tan y/2=e^c\cdot e^{-\cos x}$

$\Rightarrow e^{\cos x}\tan \frac{y}{2}=e^c=c^{\mathrm{\prime }}.$