Answer
The slope of the tangent line is $~-\pi$
Work Step by Step
$r = \frac{1}{\theta}$
$x = r~cos~\theta = \frac{cos~\theta}{\theta}$
$y = r~sin~\theta = \frac{sin~\theta}{\theta}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{\theta~cos~\theta-sin~\theta}{\theta^2}}{\frac{-\theta~sin~\theta-cos~\theta}{\theta^2}} = \frac{sin~\theta-\theta~cos~\theta}{\theta~sin~\theta+cos~\theta}$
When $\theta = \pi$:
$\frac{dy}{dx} = \frac{sin~\pi-\pi~cos~\pi}{\pi~sin~\pi+cos~\pi}$
$\frac{dy}{dx} = \frac{0-\pi~(-1)}{\pi~(0)+(-1)}$
$\frac{dy}{dx} = \frac{\pi}{-1}$
$\frac{dy}{dx} = -\pi$
The slope of the tangent line is $~-\pi$
