What is the equation of the normal line of $f(x)=tanx$ at $x=pi/6$ ?

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Answer 1 · 2015-12-06T20:15:41

$y=4/3x+(3sqrt3-2pi)/9$

The point which the tangent will intersect is $(pi/6,sqrt3/3)$ .

The slope of the tangent line is $f'(pi/6)$ .

$f(x)=tanx$

$f'(x)=sec^2x$

$f'(pi/6)=sec^2(pi/6)=1/(cos^2(pi/6))=1/(sqrt3/2)^2=1/(3/4)=4/3$

Thus, the tangent line passes through the point $(pi/6,sqrt3/3)$ and has a slope $4/3$ .

Write in point-slope form:

$y-sqrt3/3=4/3(x-pi/6)$

In slope-intercept form:

$y=4/3x+(3sqrt3-2pi)/9$

What is the equation of the normal line of f(x)=tanxf(x)=tanx at x=pi/6x=pi/6?