fx=2sin theta-sin20andy=2c cos theta-cos2 theta,theta in[0,2 pi]," then "(d^(@)y)/(dx^(2))" at "theta=pi is

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=2 cos theta-cos 2 theta and y=2 sin theta-sin 2 theta , then the value of (d^(2)y)/(dx^(2)) at theta=(pi)/(2) is -

if x =2 cos theta - cos 2 theta y = 2 sin theta - sin 2 theta Find (d^(2)y)/(dx^(2)) at theta = pi/2

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

If x=2cos theta -cos2 theta and y=2sin theta -sin2 theta , find (d^2y)/(dx^2) at theta=pi/2 .

If x =2 cos theta - cos 2 theta y = 2 sin theta - sin 2 theta Find (d^(2)y)/(dx^(2)) at theta = pi/2

If x=a(cos 2 theta+2 theta sin 2 theta) " and" y=a(sin 2 theta - 2 theta cos 2 theta) , find (d^(2)y)/(dx^(2)) " at" theta =(pi)/(8) .

If x=a(cos 2 theta+2 theta sin 2 theta) " and" y=a(sin 2 theta - 2 theta cos 2 theta) , find (d^(2)y)/(dx^(2)) " at" theta =(pi)/(8) .

If x=a (theta-sin theta) and y=a(1-cos theta) , find (d^(2)y)/(dx^(2)) at theta=pi.

If x =a cos ^(3) theta, y= a sin ^(3) theta then (d^(2) y)/( dx ^(2)) at theta = pi //4 is