Question

Asked Oct 11, 2019

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Step 1

Given equation of curve is f(x)=4secx

We know that slope of tangent to curve y=f(x) at x=a is f'(a).Therefore , slope of tangent to given curve at x=5pi/6 can be caculated as

Step 2

The y -coordinate of the point ,whose x-coordinate is 5pi/6 ,is f(5pi/6)

Step 3

Now, we use slope-point form of equation of line to find ...

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