# What is the equation of the normal line of f(x)=tanx at x=pi/6?

Dec 6, 2015

$y = \frac{4}{3} x + \frac{3 \sqrt{3} - 2 \pi}{9}$

#### Explanation:

The point which the tangent will intersect is $\left(\frac{\pi}{6} , \frac{\sqrt{3}}{3}\right)$.

The slope of the tangent line is $f ' \left(\frac{\pi}{6}\right)$.

$f \left(x\right) = \tan x$

$f ' \left(x\right) = {\sec}^{2} x$

$f ' \left(\frac{\pi}{6}\right) = {\sec}^{2} \left(\frac{\pi}{6}\right) = \frac{1}{{\cos}^{2} \left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} ^ 2 = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

Thus, the tangent line passes through the point $\left(\frac{\pi}{6} , \frac{\sqrt{3}}{3}\right)$ and has a slope $\frac{4}{3}$.

Write in point-slope form:

$y - \frac{\sqrt{3}}{3} = \frac{4}{3} \left(x - \frac{\pi}{6}\right)$

In slope-intercept form:

$y = \frac{4}{3} x + \frac{3 \sqrt{3} - 2 \pi}{9}$