Question

What is the slope of the line normal to the tangent line of `f(x) = sec^2x-sin^2x` at `x= (pi)/12`?

Answer 1

`y=(-74/11-128sqrt(3)/33)x+15/2-15/4sqrt(3)+37/66*pi+32/99*pi*sqrt(3)`

Explanation:

`f'(x)=2sec^2(x)tan(x)-2sin(x)cos(x)`

`f'(pi/12)=111/2-32sqrt(3)`
so the slope of the normal line is given by

`-1/(111/2-32sqrt(3))=-74/11-128sqrt(3)/33`
`f(pi/12)=15/8*(sqrt(3)-1)^2`