We could attempt to convert the polar equation into a rectangular equation. However, this is a bit difficult. Instead, we can find:
$$ \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{dy}{dx} $$
$$ y=r\sin{\theta} \hskip{2em} x=r\cos{\theta} $$
\( y=(1+2\sin{\theta})\sin{\theta} \)
$$ \frac{dy}{d\theta} = (1+2\sin{\theta})\cos{\theta}+\sin{\theta}(2\cos{\theta}) $$
$$ \frac{dy}{d\theta}\Big|_{\theta=0} = [1+2(0)](1)+(0)[2(1)] $$
$$ \frac{dy}{d\theta}\Big|_{\theta=0} =1 $$
\(x=(1+2\sin{\theta})\cos{\theta} \)
$$ \frac{dx}{d\theta} = (1+2\sin{\theta})(-\sin{\theta}) + \cos{\theta}(2\cos{\theta}) $$
$$ \frac{dx}{d\theta}\Big|_{\theta=0}=[1+2(0)](-0)+(1)[2(1)] $$
$$\frac{dx}{d\theta}\Big|_{\theta=0} = 2 $$
$$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \boxed{\frac{1}{2}}$$