Find dy/dx if
x=a(t+sint), y=a(1+cos t) at `t=pi/2`

Explanation

x=a(t+sint),y=a(1+cost)
`therefore` dx/dt=a(1+cost), dy/dt=-asint
`thereforedy/dx=dy/(dt/(dx/dt))=(-asint)/(a(1+cost))=-(sint)/(1+cost)`
`therefore[dy/dx]_(t=pi/2)=-("sin"pi/2)/(1+"cos"pi/2)=-1`