If `x=sin theta-theta cos theta `,`y=cos theta+theta sin theta` then `(d^(2)y)/dx ^2=`

if x=sin theta+theta cos theta,y=cos theta-theta sin theta then ((dy)/(dx))_(theta=(pi)/(2))=

If x=cos theta-cos2 theta,y=sin theta-sin2 theta then (dy)/(dx) is

If x=2cos theta -cos 2theta ,y=2sin theta -sin 2theta ,then (dy)/(dx) =

If x=2sin theta-sin2 theta and y=2cos theta-cos2 theta , theta in[0,2 pi], then (d^(2)y)/(dx^(2)) at theta=pi is

If x =a (cos theta +theta sin theta ), y =a (sin theta -theta costheta ) ,then (dy)/(dx) =

If x=2 cos theta- cos 2 theta and y=2 sin theta - sin 2 theta, then (dy)/(dx) =

If x=a(cos theta+theta sin theta),y=a(sin theta-theta cos theta), prove that (d^(2)x)/(dth eta^(2))=a(cos theta-theta sin theta),(d^(2)y)/(dth eta^(2))=a(sin theta+theta cos theta) and (d^(2)y)/(dx^(2))=(sec^(3)theta)/(a theta)

If x = sin theta - cos theta and y = sin theta + cos theta , then (dy)/(dx) =