How Do I Find Tangent and Normal Lines for Polar Equations? Given the equation $r=1+\sin(\theta)$, I am trying to compute:
(a) $\frac{dy}{dx}$
(b) the equation of the tangent and normal lines to the
curve at the indicated θ–value (which is $\frac\pi6$ in this case).
Computing $\frac{dy}{dx}$ is easy, since it involves simply plugging everything into the equation $\frac{(f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)) }{ (f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)}$.
When I used this equation, I got a $\frac{dy}{dx}$ of $\frac{\cos(\theta)(2\sin(\theta)+1) }{ \cos^2(\theta)-\sin(\theta)-\sin^2(\theta)}$.
At this point, I am stuck. I can't figure out how to compute the tangent and normal lines from here and I am getting an indeterminate form $\frac{\sqrt3} 0$ whenever I plug in the value of $\frac\pi6$.
Could someone please assist me in finding these equations? Thanks in advance!

$$\frac{dy}{dx}=\frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}
                     {\frac{dr}{d\theta} \cos \theta - r \sin \theta}$$
For $r = 1+\sin \theta, \quad \dfrac{dr}{d\theta} = \cos \theta$. So
\begin{align}
   \frac{dy}{dx}
   &=\frac{\cos \theta \sin \theta + r \cos \theta}
          {\cos^2 \theta - r \sin \theta} \\
   &=\frac{\sin \theta + r}
          {\cos \theta - r \tan \theta} \\
\end{align}
At $\theta_0 = \frac \pi 6, \quad 
    \sin \theta_0 = \frac 12, \quad 
    \cos \theta_0 = \frac{\sqrt{3}}{2},
    \quad \quad r_0 = \frac 32$
    $\quad x_0 = r_0 \cos \theta_0 = \frac{3\sqrt 3}{4},
\quad y_0 = r_0 \sin \theta_0 = \frac 34$.
So $\displaystyle m =\frac{dy}{dx}
   =\frac{\frac 12 + \frac 32}
     {\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}}
   = undefined$
In other words, the tangent line is a vertical line.
So the tangent line at $\theta_0 = \frac \pi 6$ would be
\begin{align}
   x &= x_0 \\
   r \cos \theta &= \frac{3\sqrt 3}{4} \\
   r &= \frac{3\sqrt 3}{4 \cos \theta}
\end{align}
The equation of the normal line would be computed using $y = y_0$

