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

Nov 6, 2017

$\text{Normal line slope} = \frac{- 2}{16 \sqrt{3} + 1}$

#### Explanation:

$f \left(x\right) = {\cot}^{2} x + \sin \left(x - \frac{\pi}{2}\right)$

First, find the slope of the tangent line of $f \left(x\right)$ at $x = \frac{5 \pi}{6}$ by finding the derivative of $f \left(x\right)$:
$f ' \left(x\right) = 2 \left(\cot x\right) \left(- {\csc}^{2} x\right) + \cos \left(x - \frac{\pi}{2}\right)$

$f ' \left(x\right) = \frac{- 2 \cos x}{{\sin}^{3} x} + \cos \left(x - \frac{\pi}{2}\right)$

$f ' \left(\frac{5 \pi}{6}\right) = \frac{- 2 \cos \left(\frac{5 \pi}{6}\right)}{{\sin}^{3} \left(\frac{5 \pi}{6}\right)} + \cos \left(\frac{5 \pi - 3 \pi}{6}\right)$

$= \frac{- 2 \left(- \frac{\sqrt{3}}{2}\right)}{{\left(\frac{1}{2}\right)}^{3}} + \cos \left(\frac{\pi}{3}\right)$

$= \frac{\sqrt{3}}{\frac{1}{8}} + \frac{1}{2}$

$= 8 \sqrt{3} + \frac{1}{2}$
Use common denominators to combine the two terms into one:
$f ' \left(\frac{5 \pi}{6}\right) = \frac{16 \sqrt{3} + 1}{2}$

Now that we have the slope of the tangent line at the point where $x = \frac{5 \pi}{6}$, find the slope of the normal line at this same point by taking the opposite reciprocal of this value:

$\text{Normal line slope} = \frac{- 1}{\frac{16 \sqrt{3} + 1}{2}}$

$\text{Normal line slope} = \frac{- 2}{16 \sqrt{3} + 1}$