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

Jun 23, 2018

$y = \left(- \frac{74}{11} - 128 \frac{\sqrt{3}}{33}\right) x + \frac{15}{2} - \frac{15}{4} \sqrt{3} + \frac{37}{66} \cdot \pi + \frac{32}{99} \cdot \pi \cdot \sqrt{3}$

#### Explanation:

$f ' \left(x\right) = 2 {\sec}^{2} \left(x\right) \tan \left(x\right) - 2 \sin \left(x\right) \cos \left(x\right)$

$f ' \left(\frac{\pi}{12}\right) = \frac{111}{2} - 32 \sqrt{3}$
so the slope of the normal line is given by

$- \frac{1}{\frac{111}{2} - 32 \sqrt{3}} = - \frac{74}{11} - 128 \frac{\sqrt{3}}{33}$
$f \left(\frac{\pi}{12}\right) = \frac{15}{8} \cdot {\left(\sqrt{3} - 1\right)}^{2}$