Answer
The slope of the tangent at that point is equal to:
$$\frac{2 - \sqrt 3}{1 -2\sqrt 3}$$
Work Step by Step
1. Write the formula for the slope of the tangent line:
$$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta}cos(\theta) - rsin(\theta)} $$
2. Calculate $\frac{dr}{d\theta}$ and substitute the equation into the formula:
$\frac{dr}{d\theta} = \frac{d(2 - sin(\theta))}{d\theta} = -cos(\theta)$
$\frac{dy}{dx} = \frac{-cos(\theta)sin(\theta)) + (2 - sin(\theta))cos(\theta)}{-cos(\theta)cos(\theta) - (2 - sin(\theta))sin(\theta)}$
$\frac{dy}{dx} = \frac{-cos(\theta)sin(\theta)) + 2cos(\theta) - sin(\theta)cos(\theta)}{-cos^2(\theta)- 2sin(\theta) + sin^2(\theta)}$
$\frac{dy}{dx} = \frac{-2cos(\theta)sin(\theta)) + 2cos(\theta) }{-cos^2(\theta)+ sin^2(\theta)- 2sin(\theta) }$
$\frac{dy}{dx} = \frac{-sin(2\theta) + 2cos(\theta) }{-cos(2\theta)- 2sin(\theta) }$
3. Calculate the slope when $\theta = \pi/3$
$\frac{dy}{dx} = \frac{-sin(2\frac{\pi}3) + 2cos(\pi/3) }{-cos(2\frac{\pi} 3)- 2sin(\pi/3) }$
$\frac{dy}{dx} = \frac{-\frac{\sqrt 3}2 + 2(1/2) }{-(-1/2)- 2(\frac{\sqrt 3} 2) }$
$\frac{dy}{dx} = \frac{-\frac{\sqrt 3}2 + 1 }{(1/2)-{\sqrt 3} } \times \frac{2}{2} = \frac{2 - \sqrt 3}{1 -2\sqrt 3}$
