Find an equation of the tangent line to the curve at the given point.

Question

Find an equation of the tangent line to the curve at the given point.

y = sec x,    (π/6, 23/3

Answer 1

The derivative of sec x is sec x tan x = sin x/cos²x.

pi/6 = 30° for which sin is 1/2 and cos = (1/2)√3

Therefore, the derivative is (1/2)/(3/4) = 2/3

The equation of the tangent is 2/3 = [y - (2/3)√3]/[x - (π/60] which can be "massaged" into any form you need.