91Ó°ÊÓ

From the polar coordinate system, we have \(x = r\cos\theta\) and \(y = r\sin\theta\). Given the polar curve: \(r = \frac{1}{1+\sin\theta}\), substitute the expression for \(r\) into the Cartesian coordinate formulas: $$x = \frac{\cos\theta}{1+\sin\theta}$$ $$y = \frac{\sin\theta}{1+\sin\theta}$$

Now, we need to find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\): $$\frac{dy}{d\theta} = \frac{(1+\sin\theta)\cos\theta-\sin^2\theta}{(1+\sin\theta)^2}$$ $$\frac{dx}{d\theta} = \frac{-(1+\sin\theta)\sin\theta-\cos^2\theta}{(1+\sin\theta)^2}$$ Next, find the slope of the tangent: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{(1+\sin\theta)\cos\theta-\sin^2\theta}{(1+\sin\theta)^2}}{\frac{-(1+\sin\theta)\sin\theta-\cos^2\theta}{(1+\sin\theta)^2}}$$ Simplifying this gives us: $$\frac{dy}{dx} = -\frac{(1+\sin\theta)\cos\theta-\sin^2\theta}{(1+\sin\theta)\sin\theta+\cos^2\theta}$$ Find the slope at the given point \((\frac{2}{3}, \frac{\pi}{6}):\) $$\frac{dy}{dx} \Big|_{\theta=\frac{\pi}{6}} = -\frac{(1+\frac{1}{2})\frac{\sqrt{3}}{2}-\frac{1}{4}}{(1+\frac{1}{2})\frac{1}{2}+\frac{3}{4}}$$ $$\frac{dy}{dx} \Big|_{\theta=\frac{\pi}{6}} = -\frac{3}{2}$$

With the slope at the given point and the point itself, we can write the equation of the tangent line using the point-slope form \(y-y_1=m(x-x_1)\): $$y - \frac{1}{2} = -\frac{3}{2}\left(x - \frac{2}{3}\right)$$ Simplifying the equation gives the final answer: $$y = -\frac{3}{2} x + 2$$