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#