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

Jun 2, 2018

$y = - 1 \left(6 \cdot \sqrt{6}\right) \cdot x + \frac{1}{432} \cdot \left(432 + 864 \cdot \sqrt{2} + 17 \cdot \sqrt{6} \cdot \pi\right)$

#### Explanation:

We get
$f ' \left(x\right) = - \cot \left(x\right) \cdot \csc \left(x\right) - \sec \left(x\right) \cdot \tan \left(x\right)$
so
$f ' \left(17 \cdot \frac{\pi}{12}\right) = 6 \cdot \sqrt{6}$
and the slope of the normal line is given by
${m}_{N} = - \frac{1}{6 \cdot \sqrt{6}}$
and
$f \left(\frac{17}{12} \cdot \pi\right) = 1 + 2 \sqrt{2}$
From the equation

$1 + 2 \sqrt{2} = - \frac{1}{6 \cdot \sqrt{6}} \cdot \frac{17}{12} \cdot \pi + n$
we get
$n = \frac{1}{432} \cdot \left(432 + 864 \cdot \sqrt{2} + 17 \cdot \sqrt{6} \cdot \pi\right)$