If `x = 3 cos theta - 2 cos^3 theta,y = 3 sin theta - 2 sin^3 theta`, then `dy/dx` is

Answer

`cot theta`

Explanation

`"We have, "x=3 cos theta -2 cos^(3)theta`
`"and "y=3 sin theta - 2 sin ^(3)theta`
`therefore" "(dx)/(d""theta)=-3 sin theta -2xx3 cos^(2) theta(d)/(d""theta)(cos theta)`
`=-3 sin theta + 6 cos^(2) theta sin theta`
`"And "(dy)/(d""theta)=3cos theta -2xx3sin^(2)theta(d)/(d""theta)(sin theta)`
`=3 cos theta -6 sin^(2) theta. cos theta`
`"Now, "(dy)/(dx)=(dy//d""theta)/(dx//d""theta)=(3cos theta-6 sin^(2)theta cos theta)/(-3 sin theta + 6cos^(2) theta sin theta)`
`=(3cos theta(1-2 sin^(2)theta))/(3 sin theta(-1+2cos^(2)theta))=cot theta. (cos 2theta)/(cos 2theta)=cot theta`