How do you find the equation of the tangent line to the curve #f(x)=(secx)(tanx)# at x=pi/6?

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Answer 1 · 2016-07-21T15:55:32

Equation of tangent is #10x-3sqrt3y-(5pi)/3+2sqrt3=0#

To find tangent line, we need to find the point at which tangent is drawn and its slope and then use point slope form of equation.

Former is easily available from #f(x)# as #f(pi/6)=sec(pi/6)tan(pi/6)=2/sqrt3xx1/sqrt3=2/3# Hence tangent is desired at #(pi/6,2/3)#.

For slope we need #(df)/(dx)=secx xxsec^2x+tanx xxsecxtanx#

= #sec^3x+secxtan^2x# and at #x=pi/6#, it is #(2/sqrt3)^3+2/sqrt3xx(1/sqrt3)^2=8/(3sqrt3)+2/(3sqrt3)=10/(3sqrt3)#

As equation of a line of slope #m# passing through #(x_1,y_1)# is

#(y-y_1)=m(x-x_1)# and hence equation of tangent is

#(y-2/3)=10/(3sqrt3)(x-pi/6)# or

#3sqrt3y-2sqrt3=10(x-pi/6)# or

#10x-3sqrt3y-(5pi)/3+2sqrt3=0#