Question

Find an equation of the tangent line to the curve at the given point. y=secx (pi/3,2)

Answer 1

y' = secx tanx

the reference angle of pi/3 is 60 degrees.
cos(60) = 1/2, thus, sec(60) = 2
tan(60) = √3,

so y' = 2√(3)

(pi/3, 2)

y = mx + b
2 = 2√(3) (pi/3) + b
b = 2 - (2pi√(3))/3

y = 2√(3)x + 2 - (2pi√(3))/3 <== answer

Answer 2

thank you

Answer 3

np my sleek-feathered one