# What is the equation of the line normal to  f(x)=cot^2x-x^2 at  x=pi/3?

$y - \frac{1}{3} + \setminus {\pi}^{2} / 9 = \frac{3 \setminus \sqrt{3}}{8 + 2 \setminus \pi \setminus \sqrt{3}} \left(x - \setminus \frac{\pi}{3}\right)$

#### Explanation:

Given function

$f \left(x\right) = \setminus {\cot}^{2} x - {x}^{2}$

Now, the slope of the tangent line  frac{d}{dx}f(x) is given as

$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\cot}^{2} x - {x}^{2}\right)$

$f ' \left(x\right) = 2 \setminus \cot x \left(- \setminus \cos e {c}^{2} x\right) - 2 x$

$f ' \left(x\right) = - 2 \setminus \cos e {c}^{2} x \setminus \cot x - 2 x$

Now, the slope of tangent at the point $x = \setminus \frac{\pi}{3}$

$f ' \left(x = \setminus \frac{\pi}{3}\right) = - 2 \setminus \cos e {c}^{2} \left(\setminus \frac{\pi}{3}\right) \setminus \cot \left(\setminus \frac{\pi}{3}\right) - 2 \left(\setminus \frac{\pi}{3}\right)$

$= - 2 {\left(\frac{2}{\setminus} \sqrt{3}\right)}^{2} \left(\frac{1}{\setminus} \sqrt{3}\right) - \frac{2 \setminus \pi}{3}$

$= - \frac{8}{3 \setminus \sqrt{3}} - \frac{2 \setminus \pi}{3}$

Hence, the slope of normal at the same point $x = \setminus \frac{\pi}{3}$ is

$= - \frac{1}{- \frac{8}{3 \setminus \sqrt{3}} - \frac{2 \setminus \pi}{3}}$

$= \frac{3 \setminus \sqrt{3}}{8 + 2 \setminus \pi \setminus \sqrt{3}}$

Now, the y-coordinate of point $x = \setminus \frac{\pi}{3}$ is given by substituting $x = \setminus \frac{\pi}{3}$ in the given function,

$y = f \left(x = \setminus \frac{\pi}{3}\right) = \setminus {\cot}^{2} \left(\setminus \frac{\pi}{3}\right) - {\left(\setminus \frac{\pi}{3}\right)}^{2}$

$y = \frac{1}{3} - \setminus {\pi}^{2} / 9$

Hence, the equation of normal to the given curve at $\left(\setminus \frac{\pi}{3} , \frac{1}{3} - \setminus {\pi}^{2} / 9\right)$ & having slope $\frac{3 \setminus \sqrt{3}}{8 + 2 \setminus \pi \setminus \sqrt{3}}$ is given as follows

$y - \left(\frac{1}{3} - \setminus {\pi}^{2} / 9\right) = \frac{3 \setminus \sqrt{3}}{8 + 2 \setminus \pi \setminus \sqrt{3}} \left(x - \setminus \frac{\pi}{3}\right)$

$y - \frac{1}{3} + \setminus {\pi}^{2} / 9 = \frac{3 \setminus \sqrt{3}}{8 + 2 \setminus \pi \setminus \sqrt{3}} \left(x - \setminus \frac{\pi}{3}\right)$