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

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Answer 1 · 2016-05-11T10:42:05

Hence equation of tangent is #(y-(2sqrt2-1))=(4-2sqrt2)(x-pi/8)#

#f(x)=cos2x+sin2x+tanx# and

at #x=pi/8#, #f(x)=cos(pi/4)+sin(pi/4)+tan(pi/8)=1/sqrt2+1/sqrt2+sqrt2-1#

= #2/sqrt2+sqrt2-1=2sqrt2-1#

Hence tangent line passes through #(pi/8,(2sqrt2-1))#

#(df)/dx# gives slope of tangent and as

#(df)/dx=-2sin2x+2cos2x+sec^2x# and at #x=pi/8#

slope is #-2/sqrt2+2/sqrt2+(sqrt(4-2sqrt2))^2#

or #-2/sqrt2+2/sqrt2+(4-2sqrt2)=4-2sqrt2#

Hence equation of tangent is

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