Question

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

Answer 1

To get the equation of tangent, we need a point on it and its slope.

we already have (x1,y1) = (pi/3,2) on the tangent line.

To get the slope of tangent,
m = derivative of y at the given point.

y = sec x
y' = sec x tan x
at x = pi/3,
y' = sec pi/3 tan pi/3
y'= 2 $\sqrt 3$

so the slope of tangent is 2 $\sqrt 3$

equation will be
y -y1 = m (x-x1)
y - 2 = 2 $\sqrt 3$ (x- pi/3)

Thats your equation of tangent.
you can simplify if you want.