Answer
The equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=-\frac{7}{6} t+\frac{\sqrt{3}}{6}+\frac{ 7\pi}{18} .$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\cos x)'=-\sin x$.
Recall that $(\cot x)'=-\csc^2 x$.
Since $ y=\cot t\cos t $, then using the product rule one can find that $ y'=-\csc^2 t \cos t-\cot t \sin t $ and hence the slope at $ t=\frac{\pi}{3}$ is $ m=-\csc^2 \frac{\pi}{3} \cos \frac{\pi}{3}-\cot \frac{\pi}{3} \sin \frac{\pi}{3}=-\frac{7}{6}.$ Now, the equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=-\frac{7}{6} t+c.$$
Since the curve and the line coincide at $ t=\frac{\pi}{3}$, then
$$\cot \frac{\pi}{3} \cos \frac{\pi}{3}=-\frac{7}{6} t+c \Longrightarrow c= \frac{\sqrt{3}}{6}+\frac{ 7\pi}{18} .$$
Hence, the equation of the tangent line at $ t=\frac{\pi}{3}$ is given by
$$ y=-\frac{7}{6} t+\frac{\sqrt{3}}{6}+\frac{ 7\pi}{18} .$$
