Answer
$TLS\vert_{\frac{\pi}{4}}=1$
Work Step by Step
Given:
$r=cos(2(\frac{\pi}{4}))$
$\theta=\frac{\pi}{4}$
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 $\frac{dr}{d\theta}$:
$\frac{dr}{d\theta}=(cos(2(\frac{\pi}{4})))^{\prime}={-2sin(2(\frac{\pi}{4}))}$
Plug in for $r$, $\theta$, and $\frac{dr}{d\theta}$:
$TLS\vert_{\frac{\pi}{4}}=\frac{(-2sin(2(\frac{\pi}{4})))sin(\frac{\pi}{4})+(cos(2(\frac{\pi}{4})))cos(\frac{\pi}{4})}{({-2sin(2(\frac{\pi}{4}))})cos(\frac{\pi}{4})-(cos(2(\frac{\pi}{4})))sin(\frac{\pi}{4})}=\frac{(-2sin(2(\frac{\pi}{4})))sin(\frac{\pi}{4})+(cos(2(\frac{\pi}{4})))cos(\frac{\pi}{4})}{({-2sin(2(\frac{\pi}{4}))})cos(\frac{\pi}{4})-(cos(2(\frac{\pi}{4})))sin(\frac{\pi}{4})}$
Solve:
$sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$
$cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$
$sin(2*\frac{\pi}{4})=1$
$cos(2*\frac{\pi}{4})=0$
Simplify:
$TLS\vert_{\frac{\pi}{4}}=\frac{-2(1)(\frac{1}{\sqrt2})+(0)(\frac{1}{\sqrt2})}{{-2(1)(\frac{1}{\sqrt2})}-(0)(\frac{1}{\sqrt2})}=\frac{-2(1)(\frac{1}{\sqrt2})}{{-2(1)(\frac{1}{\sqrt2})}}=1$
