Locus of centroid of triangle formed by `A(2cos theta,3sin theta),B(-cos theta,0),C(0,2sin theta)` is (where `theta` is variable )

(cos^(2)theta)/(sin theta)-cosec theta+sin theta=0

The locus of the point (a(cos theta+sin theta),b(cos theta-sin theta)) where theta is the parameter is

cos^(2)theta-sin theta.cos theta-(1)/(2)=0

cos^(2)theta-sin theta*cos theta-(1)/(2)=0

cos^(2)theta-sin theta*cos theta-(1)/(2)=0

cos^(2)theta-sin theta*cos theta-(1)/(2)=0

cos^(2)theta-sin theta*cos theta-(1)/(2)=0

(sin3 theta+cos3 theta)/(cos theta-sin theta)=1+2xx sin2 theta

Show that the centroid of the triangle with vertices (5 cos theta, 4 sin theta), (4 cos theta, 5 sin theta) and (0, 0) lies on the circle x^(2)+y^(2)=9 .

If sin2 theta-cos3 theta=0 and theta is acute then theta =