find the line slope of the tangent line to the given polar?

Question

find the line slope of the tangent line to the given polar curve at the point specified by the value of θ
r=2-sin(θ)
θ= pi/3

ok i know that i have take the derivative (dy/dx)
I started off with
y=rsin(θ)=(2-sin(θ))(sin(θ)=2sin(θ)-(sinθ)^2
x=rcos(θ)=(2-sinθ)(sinθ)= 2cosθ-sinθcosθ
i got jumbled somewhere and i didn't get the right answer
which is (2-sqrt3)/(1-2sqrt3)
thanks for the help.

Answer 1

Find the line slope of the tangent line to the given polar curve at the point specified by the value of θ.

r = 2 - sinθ
θ = π/3

You had the right idea.

Remember the identities and plug in for r.

y = rsinθ = (2 - sinθ)sinθ = 2sinθ - sin²θ
x = rcosθ = (2 - sinθ)cosθ = 2cosθ - sinθcosθ

Take the derivative of y and x with respect to θ.

dy/dθ = 2cosθ - 2sinθcosθ
dx/dθ = -2sinθ - cos²θ + sin²θ

Solve for dy/dx.

dy/dx = (dy/dθ) / (dx/dθ)

dy/dx = (2cosθ - 2sinθcosθ) / (-2sinθ - cos²θ + sin²θ)

Plug in the values for θ = π/3.

dy/dx = [2(1/2) - 2(√3/2)(1/2)] / [-2(√3/2) - 1/4 + 3/4]

dy/dx = [1 - √3/2] / [-√3 + 1/2] = (2 - √3) / (-2√3 + 1)

dy/dx = (2 - √3) / (1 -2√3)

Answer 2

use a scientific calculator and do a polar to rectangular conversion