If `x=3costheta-cos^3theta` `y=3sintheta-sin^3theta` find `dy/dx`

Answer

N/a

Explanation

`:' x=3costheta-2cos^(3)theta` and ` y = 3sin theta - 2 sin^(3) theta`
`:. (dx)/(dy) = (d)/(dx) (3cos theta)-(d)/(d theta)(2cos^(3) theta)`
`= 3.(-sin theta)-2.3 cos^(2) theta.(d)/(d theta). cos theta`
`= -3 sin theta+ 6 cos^(2) theta sin theta`
and `(dy)/(d theta) = 3 cos A - 2.3 sin^(2) theta.(d)/(d theta) . sin theta`
`= 3 cos theta - 6 sin^(2) theta.costheta`
Now, `(dy)/(dx) = (dy//d theta)/(dx//d theta)= (3costheta-6sin^(2) thetacostheta)/(-3sin theta+6cos^(2) thetasin theta)`
`= (3costheta(1-2sin^(2) theta))/(3sin theta(-1+2cos^(2) theta)) = cot theta. (cos2 theta)/(cos 2 theta) = cot theta`