What is the equation of the normal line of #f(x)= tan(2x)/cotx# at #x = pi/8#?

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Answer 1 · 2016-09-23T11:54:09

#-2(sqrt2+1)(2-sqrt2)(y-(2-sqrt2)/2)=x-pi/8#

equation of the normal: #-y'_0(y-y_0)=x-x_0#

#y'=((2/cos^2(2x))*cotx+(tan2x)/sin^2x)/cot^2x#

#y'_0=((2/cos^2(pi/4))*cot(pi/8)+(tan(pi/4))/sin^2(pi/8))/cot^2(pi/8)=#

#=(4cos(pi/8)sin(pi/8)+1)/cos^2(pi/8)=#

#=(2sin(pi/4)+1)/((cos(pi/4)+1)/2)=#

#=4*(sqrt2+1)/(sqrt2+2)#

#y_0=tan(pi/8)=1-sqrt2/2#

#-2(sqrt2+1)(2-sqrt2)(y-(2-sqrt2)/2)=x-pi/8#