Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 4


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

Work Step by Step

Since $ y=\sec x $, then $ y'=\sec x \tan x $ and hence the slope at $ x=\frac{\pi}{6}$ is $ m=\sec \frac{\pi}{6}\tan \frac{\pi}{6}=\frac{2}{3}.$ Now, the tangent line equation is given by $$ y=\frac{2}{3} x+c.$$ Since the curve and line coincide at $ x=\frac{\pi}{6}$, then $$\sec \frac{\pi}{6}=\frac{2}{3}\frac{\pi}{6}+c \Longrightarrow c= \frac{2}{\sqrt{3}}-\frac{ \pi}{9} .$$ Hence the equation of the tangent line is given by $$ y=\frac{2}{3} x+\frac{2}{\sqrt{3}}-\frac{ \pi}{9} .$$
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