If `x = 3 cos t - 2 cos^3 t and y = 3 sin t - 2 sin^3 t`, show that `dy/dx = cott`

If x = 2 cos t - cos 2t , y = 2 sin t - sin 2t , then the value of d^2y/dx^2 at t=pi//2 is

Show that the equation of normal at any point 't' on the curve: x = 3 cos t - cos^3 t and y = 3 sint - sin^3 t is: 4(ycos^3t - x sin^3t)=3 sin 4t .

If x = sin t, y = cos t, find dy/dx

x= a sin 2t(1 + cos2t) and y = b cos 2t(1- cos 2t) , show that [(dy/dx)_(t=pi/4) =b/a] .

If x = a sin 2 t(1+cos 2t) and y = b cos 2t(1-cos 2t), find: the value of dy/dx at t = pi/4 and t = pi/3

If x=cost(3-2cos^(2)t) and y=sint(3-2sin^(2)t) , find the value of (dy)/(dx) at t=(pi)/(4) .

Find dy/dx at t = pi/4 , when x = 2 cos 2t and y = 2 sin t - sin 2t

Find dy/dx when x= a cos^3 t , y = b sin^3 t

If x^2 + y^2 = t + 1/t and x^4 + y^4 = t^2 + 1/t^2 , then show that dy/dx = -1/(x^3y)

If x = (1 - t^2)/(1 + t^2) and y = (2t)/(1+t^2) , then show that (dy/dx) + x/y =0