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

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Student Comments

balajia | Student

The slope of the tangent is equal to the derivative of the equation of the curve at the given point.

`dy/dx = secx.tanx`

The slope of the tangent is equal to `2sqrt3.`

The equation of the tangent is equal to `y-2=2sqrt3(x-pi/3)`

that is `2sqrt3x-y=(2pi)/sqrt3-2`

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