Find an equation of the tangent line to the curve y = sec x at the at the point `(pi/6, 2/sqrt3)`

1 Answer | Add Yours

justaguide's profile pic

Posted on

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`

We’ve answered 324,509 questions. We can answer yours, too.

Ask a question