Answer
\[\frac{\sqrt 3}{9}\]
Work Step by Step
\[r=1+2\cos\theta\]
Differentiate $r$ with respect to $\theta$:
\[\frac{dr}{d\theta}=-2\sin\theta\]
Slope of tangent line to the polar curve is given by:
\[m=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}\]
\[m=\frac{(-2\sin\theta)(\sin\theta)+(1+2\cos\theta)\cos\theta}{(-2\sin\theta)\cos\theta-(1+2\cos\theta)\sin\theta}\]
at $\theta=\displaystyle\frac{\pi}{3}$
\[m=\frac{\left[-2\times\displaystyle\frac{\sqrt{3}}{2}\right]\displaystyle\frac{\sqrt{3}}{2}+\left[1+2\times\displaystyle\frac{1}{2}\right]\displaystyle\frac{1}{2}}{\left[-2\times\displaystyle\frac{\sqrt{3}}{2}\right]\displaystyle\frac{1}{2}-\left[1+2\times\displaystyle\frac{1}{2}\right]\displaystyle\frac{\sqrt{3}}{2}}\]
\[\Rightarrow m=\frac{(-\sqrt{3})\left(\displaystyle\frac{\sqrt 3}{2}\right)+(1+1)\displaystyle\frac{1}{2}}{-\displaystyle\frac{\sqrt{3}}{2}-\sqrt{3}}\]
\[\Rightarrow m=\frac{-\displaystyle\frac{3}{2}+1}{\displaystyle\frac{-3\sqrt{3}}{2}}\]
\[\Rightarrow m=\frac{-1}{3\sqrt{3}}=\frac{1}{3\sqrt{3}}=\frac{\sqrt 3}{9}\]
