What is the equation of the normal line of #f(x)= tanx# at #x = pi/8#?

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Answer 1 · 2017-11-12T12:47:21

Equation of normal line is # x+ 1.17y =0.87#

#x=pi/8 ~~0.39 ; f(x)=tanx = or f(x)= tan (pi/8) ~~ 0.41#

So at #(0.39,0.41)# tangent and normal is drawn.

Slope of the tangent is #f'(x)= sec^2x#. at #x=pi/8#

#sec^2x=1/cos^2x= 1/(cos (pi/8))^2 ~~ 1.17# . Slope of tangent

is #m_1=1.17# Normal line is perpendicular to tangent , so

slope of the normal line # m_2= -1/m_1 = -1/1.17#

Equation of normal line at point #(0.39,0.41) ; m_2= -1/1.17#

is # y-y_1=m_2(x-x_1) or (y- 0.41) = -1/1.17(x-0.39)# or

# 1.17 (y- 0.41) = -(x-0.39) # or

#x+1.17y=0.39+(1.17*0.41)# or

# x+1.17y=0.39+(1.17*0.41) or x+1.17y =0.87#

Equation of normal line is # x+ 1.17y =0.87# [Ans]