Answer
The equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=\frac{2}{3} t+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$
Work Step by Step
Since $ y=\frac{\sin t}{1+\cos t}$, then using the quotient rule one can find that $ y'=\frac{1}{1+\cos t}$ and hence the slope at $ t=\frac{\pi}{3}$ is $ m=\frac{1}{1+\cos \frac{\pi}{3}}=\frac{2}{3}.$ Now, the equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=\frac{2}{3} t+c.$$
Since the curve and line coincide at $ t=\frac{\pi}{3}$, then
$$\frac{\sin \frac{\pi}{3}}{1+\cos \frac{\pi}{3}}=\frac{2}{3}\frac{\pi}{3}+c \Longrightarrow c= \frac{1}{\sqrt{3}}-\frac{ 2\pi}{9} .$$
Hence the equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=\frac{2}{3} t+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$
