# What is the equation of the line normal to  f(x)=sec4x-cot2x at  x=pi/3?

Apr 11, 2018

$\text{Normal} \implies y = - \frac{3 x}{8 - 24 \sqrt{3}} + \frac{152 \sqrt{3} - 120 + 3 \pi}{24 - 72 \sqrt{2}} \implies y \approx 0.089 x - 1.52$

#### Explanation:

The normal is the perpendicular line to the tangent.

$f \left(x\right) = \sec \left(4 x\right) - \cot \left(2 x\right)$
$f ' \left(x\right) = 4 \sec \left(4 x\right) \tan \left(3 x\right) + 2 {\csc}^{2} \left(2 x\right)$

$f ' \left(\frac{\pi}{3}\right) = 4 \sec \left(\frac{4 \pi}{3}\right) \tan \left(\frac{4 \pi}{3}\right) + 2 {\csc}^{2} \left(\frac{2 \pi}{3}\right) = \frac{8 - 24 \sqrt{3}}{3}$

For normal, $m = - \frac{1}{f ' \left(\frac{\pi}{3}\right)} = - \frac{3}{8 - 24 \sqrt{3}}$

$f \left(\frac{\pi}{3}\right) = \sec \left(\frac{4 \pi}{3}\right) - \cot \left(\frac{2 \pi}{3}\right) = \frac{\sqrt{3} - 6}{3}$

$\frac{\sqrt{3} - 6}{3} = - \frac{3}{8 - 24 \sqrt{3}} \left(\frac{\pi}{3}\right) + c$

$c = \frac{\sqrt{3} - 6}{3} + \frac{\pi}{8 - 24 \sqrt{3}} = \frac{152 \sqrt{3} - 120 + 3 \pi}{24 - 72 \sqrt{2}}$

"Normal":y=-(3x)/(8-24sqrt3)+(152sqrt3-120+3pi)/(24-72sqrt2);y=0.089x-1.52