Question

What is the slope of the line normal to the tangent line of `f(x) = cscx-sec^2x +1` at `x= (17pi)/12`?

Answer 1

`y=-1(6*sqrt(6))*x+1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)`

Explanation:

We get
`f'(x)=-cot(x)*csc(x)-sec(x)*tan(x)`
so
`f'(17*pi/12)=6*sqrt(6)`
and the slope of the normal line is given by
`m_N=-1/(6*sqrt(6))`
and
`f(17/12*pi)=1+2sqrt(2)`
From the equation

`1+2sqrt(2)=-1/(6*sqrt(6))*17/12*pi+n`
we get
`n=1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)`