If $\frac{3}{2+\cos \theta +i\sin \theta }=x+iy$, then $(x-1)(x-3)=$

$\frac{3}{2+\cos \theta +i\sin \theta }=x+iy$

$\frac{3}{(2+\cos \theta )+\sin \theta }\times \frac{(2+\cos \theta )-i\sin \theta }{(2+\cos \theta )-i\sin \theta }=x+iy$

$\frac{6+3\cos \theta -3i\sin \theta }{5+4\cos \theta }=x+iy$

$x=\frac{6+3\cos \theta }{5+4\cos \theta },y=\frac{-3\sin \theta }{5+4\cos \theta }$

$(x-1)(x-3)=\frac{-9(1-{\cos }^2\theta )}{{(5+4\cos \theta )}^2}$

$(x-1)(x-3)=-y^2$