What is the equation of the tangent line of #f(x) =(sin2x)/(cos2x)-tanx# at #x=pi/8#?

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Answer 1 · 2016-03-30T11:12:55

#8x-2sqrt2y+pi+4-4sqrt2=0#

#f(x)=(sin2x)/(cos2x)-tanx=tan2x-tanx#

At #x=pi/8#, #f(x)=tan(pi/8xx2)-tanpi/8#

= #tan(pi/4)-tan(pi/8)=1-(sqrt2-1)=2-sqrt2#

Hence tangent touches curve at point #(pi/8,2-sqrt2)#

#f'(x)=2sec^2(2x)-sec^2x# and at #x=pi/8#

#f'(x)=2sec^2(pi/4)-sec^2(pi/8)=2(sqrt2)^2-(4-2sqrt2)#

= #4-4+2sqrt2=2sqrt2#

This gives the slope of the tangent.

Hence, using point slope form, equation of tangent is

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

#y-2+sqrt2=2sqrt2x-pi/(2sqrt2)# or

#2sqrt2y-4sqrt2+4=8x-pi# or

#8x-2sqrt2y+pi+4-4sqrt2=0#