Find the equation of tangent to the curve #y=secx-tanx# at point #x=0#?

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Answer 1 · 2017-12-19T09:50:33

Equation of tangent is #x+y=1#

The slope of a tangent at #x=x_0# to a function #y=f(x)# is given by #f'(x_0)#, where #f'(x)=(df)/(dx)#

Here we are seeking tangent to #y=secx-tanx# at #(0,1)#

First checking it #->#as #x=0# we have #y=sec0-tan0=1-0=1#

and hence #(1,0)# indeed lies on the curve #y=secx-tanx#

Now as #y=secx-tanx#, #f'(x)=secxtanx-sec^2x#

and slope of tangent at #(0,1)# is

#sec0tan0-sec^2 0=1*0-1^2=-1#

and hence using point slope form equation of tangent is

#y-1=-1(x-0)# or #x+y=1#

graph{(y-secx+tanx)(x+y-1)(x^2+(y-1)^2-0.01)=0 [-4.98, 5.02, -1.66, 3.34]}