`x=(sin^(3)t)/(sqrt(cos 2t))`

If x and y are connected parametrically by the equations given, without eliminating the parameter, Find (dy)/(dx) . x=(sin^3t)/(sqrt(cos2t)), y=(cos^3t)/(sqrt(cos2t))

x=sqrt(sin 2t),y=sqrt(cos 2 t) find dy/dx

If x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t) show that dy/dx =0 at t=pi/6

Solve l=int_(cos^(4)t)^(-sin^(4)t)(sqrt(f(z))dz)/(sqrt(f(cos 2 t -z))+sqrt(f(z)))

The graph of {:{(x=sin^(2)t),(y=2 cos t):}

If x=sqrt(a^(sin^(-1)t)),y=sqrt(a^(cos^(-1)t) , a >0 a n d -1 < t < 1 , show that (dy)/(dx)=-y/x .

Find (dy)/(dx), when x=cos^(-1)1/(sqrt(1+t^2))"and"y=sin^(-1)1/(sqrt(1+t^2)),tR

If y = cos^(-1) ((1)/( sqrt(1+t^(2)))), x = sin^(-1) (sqrt((t^(2))/(1 + t^(2)))), "find " (dy)/(dx)

If x=sqrt(a^(sin^(-1)t)), y=sqrt(a^(cos^(-1)t)) , Show that (dy)/(dx)=-y/x .

If x=sqrt(a^(sin^(-1)t)), y=sqrt(a^(cos^(-1)t)) , show that (dy)/(dx)=-y/x