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

y = sec x, (π/6, 2√3/3)

The equation of tangent to the curve f(x) at (a , b) can be found by finding the slope of tangent which is f^{'}(a)

f(x) = sec x

f^{'}(x)= secx tanx

Now at the point (π/6, 2√3/3)

f^{'}(x) = sec π/6 tan π/6

f^{'}(x) = (2/√3)(1/√3)

f^{'}(x) = 2/3

Slope = m

Thus the required equation is (y - 2√3/3) = 2/3 (x - π/6).

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

y = sec x, (π/6, 2√3/3)

**Summary:**

The equation of the tangent line to the curve at the given point is (y - 2√3/3) = 2/3(x - π/6).

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