Answer
$TLS\vert_{{\pi}}=-\pi$
Work Step by Step
Given:
$r=\frac{1}{{{\pi}}}$
$\theta={\pi}$
Use the equation for tangent line slope for polar coordinates:
$TLS=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$
Find $\frac{dr}{d\theta}$:
$\frac{dr}{d\theta}=(\frac{1}{{{\pi}}})^{\prime}={-\frac{1}{{{\pi}}^2}}$
Plug in for $r$, $\theta$, and $\frac{dr}{d\theta}$:
$TLS\vert_{\pi}=\frac{(-\frac{1}{{{\pi}}^2})sin({\pi})+(\frac{1}{{{\pi}}})cos({\pi})}{({-\frac{1}{{{\pi}}^2}})cos({\pi})-(\frac{1}{{{\pi}}})sin({\pi})}=\frac{(-\frac{1}{{{\pi}}^2})sin({\pi})+(\frac{1}{{{\pi}}})cos({\pi})}{({-\frac{1}{{{\pi}}^2}})cos({\pi})-(\frac{1}{{{\pi}}})sin({\pi})}$
Solve:
$sin({\pi})=0$
$cos({\pi})=-1$
Simplify:
$TLS\vert_{{\pi}}=\frac{(-\frac{1}{{{\pi}}^2})(0)+(\frac{1}{{{\pi}}})(1)}{({-\frac{1}{{{\pi}}^2}})(1)-(\frac{1}{{{\pi}}})(0)}=\frac{\frac{1}{\pi}}{-\frac{1}{\pi^2}}=\frac{1}{\pi}*-\frac{\pi^2}{1}=-\pi$
