Find `(dy)/(dx),` when `x=a(theta+sintheta)a n dy=a(1-costheta)`

Explanation

`x = a(theta+sintheta)`

`:. dx/(d theta) = a(1+cos theta)`

`y = a(1-cos theta)`

`:. dy/( d theta) = asin theta`

`:. dy/dx = (dy/( d theta))/(dx/( d theta)) = ( asin theta)/(a(1+cos theta))`

`=>dy/dx = sin theta/(1+cos theta) = (2sin(theta/2)cos (theta/2))/(2cos^2(theta/2)) = tan(theta/2)`

`=>:. dy/dx = tan(theta/2)`.