Answer
$TLS\vert_{\frac{\pi}{4}}=\frac{1-\sqrt2}{2+\sqrt2}$
Work Step by Step
Given:
$r=2+sin(3(\frac{\pi}{4}))$
$\theta=\frac{\pi}{4}$
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 $\frac{dr}{d\theta}$:
$\frac{dr}{d\theta}=(2+sin(3(\frac{\pi}{4})))^{\prime}={3cos(3(\frac{\pi}{4}))}$
Plug in for $r$, $\theta$, and $\frac{dr}{d\theta}$:
$TLS\vert_{\frac{\pi}{4}}=\frac{(3cos(3(\frac{\pi}{4})))sin(\frac{\pi}{4})+(2+sin(3(\frac{\pi}{4})))cos(\frac{\pi}{4})}{({3cos(3(\frac{\pi}{4}))})cos(\frac{\pi}{4})-(2+sin(3(\frac{\pi}{4})))sin(\frac{\pi}{4})}=\frac{(3cos(3(\frac{\pi}{4})))sin(\frac{\pi}{4})+(2+sin(3(\frac{\pi}{4})))cos(\frac{\pi}{4})}{({3cos(3(\frac{\pi}{4}))})cos(\frac{\pi}{4})-(2+sin(3(\frac{\pi}{4})))sin(\frac{\pi}{4})}$
Solve:
$sin(\frac{3\pi}{4})=\frac{1}{\sqrt{2}}$
$cos(\frac{3\pi}{4})=-\frac{1}{\sqrt{2}}$
$sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$
$cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$
Simplify:
$TLS\vert_{\frac{\pi}{4}}=\frac{3(\frac{-1}{\sqrt{2}})(\frac{1}{\sqrt{2}})+2(\frac{1}{\sqrt{2}})+(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})}{3(\frac{-1}{\sqrt{2}})(\frac{1}{\sqrt{2}})-2(\frac{1}{\sqrt{2}})-(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})}=\frac{-3(\frac{1}{2})+2(\frac{1}{2})+(\frac{1}{2})}{-3(\frac{1}{2})-2(\frac{1}{\sqrt2})-(\frac{1}{2})}=\frac{-1+\frac{2\sqrt2}{2}}{-2-(\frac{2\sqrt2}{2})}=\frac{1-\sqrt2}{2+\sqrt2}$
