Find the points at which the polar curve $r=2+2\sin{(\theta)}$ has a horizontal or vertical tangent line.
Translate the parametric equation to Cartesian coordinates: $$ r^2=2r+2r\sin{(\theta)} \\ x^2+y^2=2\sqrt{x^2+y^2}+2y \\ x^2+y^2-2y+1=2\sqrt{x^2+y^2}+1 \\ x^2+(y-1)^2=2\sqrt{x^2+y^2}+1 \\ $$ Then differentiate with respect to $x$: $$ \frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\ 2x+2(y-1)y'=\frac{2(x+y)}{\sqrt{x^2+y^2}} \\ x+(y-1)y'=\frac{x+y}{\sqrt{x^2+y^2}}y' \\ y'=\frac{x}{\frac{x+y}{\sqrt{x^2+y^2}}-(y-1)} \\ y'=\frac{x}{\frac{x+y-\sqrt{x^2+y^2}(y-1)}{\sqrt{x^2+y^2}}} \\ y'=\frac{x\sqrt{x^2+y^2}}{x+y-\sqrt{x^2+y^2}(y-1)} \\ $$
So, by now it seems I'm not making any real progress in finding the vertical and horizontal tangents of this equation.
Am I going about this wrong?
Mistake in taking derivatives: $$ \frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right] =\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\ 2x+2(y-1)y'=\frac{2x+2yy'}{\sqrt{x^2+y^2}} $$ and you dropped the $y'$ on the RHS...
UPDATE Another approach is to let $x=x(\theta)$ and $y = y(\theta)$ and compute $$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} $$ In our case, $r=2+2\sin \theta$ so $$x = r\cos \theta = \cos \theta (2+2\sin \theta)$$ and similarly $$y = r\sin \theta = \sin \theta (2+2\sin \theta)$$, so computing $y'(\theta)$ is straight-forward arithmetic.
Setting it to $0$ and $\pm \infty$ should give you the relevant angles...