Question

What is the slope of the line normal to the tangent line of `f(x) = cot^2x+sin(x-pi/4)` at `x= (5pi)/6`?

Answer 1

the slope of the normal line is given by
`1/586753(sqrt(6)+sqrt(2)-32*sqrt(3))*(749+48*sqrt(2))`

Explanation:

`f'(x)=cos(pi/4-x)-2cot(x)*csc^2(x)`
so
`f'(5*pi/6)=1/4*(sqrt(2)+32sqrt(3)-sqrt(6))`
so the slope of the normal line is given by
`m_N=-1/(1/4*(sqrt(2)+32*sqrt(3)-sqrt(6))`