# 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 ?

Jun 2, 2018

the slope of the normal line is given by
$\frac{1}{586753} \left(\sqrt{6} + \sqrt{2} - 32 \cdot \sqrt{3}\right) \cdot \left(749 + 48 \cdot \sqrt{2}\right)$

#### Explanation:

$f ' \left(x\right) = \cos \left(\frac{\pi}{4} - x\right) - 2 \cot \left(x\right) \cdot {\csc}^{2} \left(x\right)$
so
$f ' \left(5 \cdot \frac{\pi}{6}\right) = \frac{1}{4} \cdot \left(\sqrt{2} + 32 \sqrt{3} - \sqrt{6}\right)$
so the slope of the normal line is given by
m_N=-1/(1/4*(sqrt(2)+32*sqrt(3)-sqrt(6))