# What is the equation of the line that is normal to f(x)=cscx+tanx-cotx  at  x=-pi/3?

Feb 23, 2018

$y = - \frac{3 x}{14} - 2.53$

#### Explanation:

$\text{Tangent} : \frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = f ' \left(x\right)$

$\text{Normal} : - \frac{1}{f ' \left(x\right)} = - \frac{1}{\frac{d}{\mathrm{dx}} \left[\csc x + \tan x - \cot x\right]} = - \frac{1}{\frac{d}{\mathrm{dx}} \left[\csc x\right] + \frac{d}{\mathrm{dx}} \left[\tan x\right] - \frac{d}{\mathrm{dx}} \left[\cot x\right]} = - \frac{1}{- \csc x \cot x + {\sec}^{2} x + {\csc}^{2} x}$

$- \frac{1}{f ' \left(- \frac{\pi}{3}\right)} = - \frac{1}{- \csc \left(- \frac{\pi}{3}\right) \cot \left(- \frac{\pi}{3}\right) + {\sec}^{2} \left(- \frac{\pi}{3}\right) + {\csc}^{2} \left(- \frac{\pi}{3}\right)} = - \frac{1}{\frac{14}{3}} = - \frac{3}{14}$

$y = m x + c$

$f \left(a\right) = m a + c$

$\csc \left(- \frac{\pi}{3}\right) + \tan \left(- \frac{\pi}{3}\right) - \cot \left(- \frac{\pi}{3}\right) = - \frac{\pi}{3} \left(- \frac{3}{14}\right) + c$

$c = \csc \left(- \frac{\pi}{3}\right) + \tan \left(- \frac{\pi}{3}\right) - \cot \left(- \frac{\pi}{3}\right) + \frac{\pi}{3} \left(- \frac{3}{14}\right)$

$c = - 2.53$

$y = - \frac{3 x}{14} - 2.53$