# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 27

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 $$y=\frac{2}{3} t+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$

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