Find an equation of the tangent line to the curve y = sec x at the at the point `(pi/6, 2/sqrt3)`
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The equation of the tangent to the curve y = sec x at the point `(pi/6, 2/sqrt 3)` has to be determined.
y' = sec x*tan x
The slope of the line tangent to y = sec x at any point is equal to the value of sec x*tan x at that point.
y' at `pi/6` = `sec (pi/6)*tan (pi/6)` = `2/3`
The equation of the tangent is `(y - 2/sqrt 3)/(x - pi/6) = 2/3`
=> `3y - 6/sqrt 3 = 2x - pi/3`
=> `2x - 3y - pi/3 + 2*sqrt 3 = 0`
The required tangent is `2x - 3y - pi/3 + 2*sqrt 3 = 0`
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