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

Jul 26, 2016

$3 x - 6 y + \pi - 6 \sqrt{3} = 0$.

#### Explanation:

We know that $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$ gives the slope of tgt. to curve $C : y = f \left(x\right)$ at any pt.$P \left(x , y\right)$ on the curve.

since, normal line is $\bot$ to tgt., its slope, at any pt. $P \left(x , y\right)$, will be $- \frac{1}{f ' \left(x\right)} , \mathmr{if} f ' \left(x\right) \ne 0$

The Curve is $C : y = f \left(x\right) = \csc x + \cot x$

$\therefore f ' \left(x\right) = - \csc x \cot x - {\csc}^{2} x = - \csc x \left(\cot x + \csc x\right)$

$\therefore f ' \left(- \frac{\pi}{3}\right) = - \csc \left(- \frac{\pi}{3}\right) \left\{\cot \left(- \frac{\pi}{3}\right) + \csc \left(- \frac{\pi}{3}\right)\right\}$

$= - \frac{2}{\sqrt{3}} \left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right) = - \frac{6}{3} = - 2$

Therefore, the slope of normal $= \frac{1}{2}$.

Also, $f \left(- \frac{\pi}{3}\right) = \csc \left(- \frac{\pi}{3}\right) + \cot \left(- \frac{\pi}{3}\right) = - \sqrt{3}$.

Thus, the normal line passes thro. pt. $\left(- \frac{\pi}{3} , - \sqrt{3}\right)$ and has slope$= \frac{1}{2}$.

Therefore, the eqn. of normal is given by,

$y + \sqrt{3} = \frac{1}{2} \left(x + \frac{\pi}{3}\right)$, i.e., $6 y + 6 \sqrt{3} = 3 x + \pi$, or

$3 x - 6 y + \pi - 6 \sqrt{3} = 0$.