Solution of $2 y \sin x \frac{d y}{d x}=2 \sin x \cos x-y^2 \cos x, x=\frac{\pi}{2}, y=1$ is given by

$y^2=\sin x$

On dividing by sin x,

$2 y \frac{d y}{d x}+y^2 \cot x=2 \cos x$

Put $y^2=v \Rightarrow \frac{d v}{d x}+v \cot x=2 \cos x$

I.F. $=e^{\int \cot x d x}=e^{\log \sin x}=\sin x$ 

∴ Solution is $v . \sin x \int \sin x .(2 \cos x) dx+c$

$\Rightarrow y^2 \sin x=\sin ^2 x+c$

When $x=\frac{\pi}{2}$, y = 1, then c = 0

∴ $y^2=\sin x$

Hence (1) is the correct answer.