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

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May 12, 2018

The equation of normal is $3 x + 15 y = 28.14$

Explanation:

$\sec \left(\frac{\pi}{3}\right) = 2 , \tan \left(\frac{\pi}{3}\right) = \sqrt{3} , \cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$

${\cot}^{2} \left(\frac{\pi}{3}\right) = \frac{1}{3} , \csc \left(\frac{\pi}{3}\right) = \frac{2}{\sqrt{3}} , {\csc}^{2} \left(\frac{\pi}{3}\right) = \frac{4}{3} ,$

$f \left(x\right) = \sec x - {\cot}^{2} x \therefore f \left(\frac{\pi}{4}\right) = \sec \left(\frac{\pi}{4}\right) - {\cot}^{2} \left(\frac{\pi}{4}\right)$ or

$f \left(\frac{\pi}{4}\right) = 2 - \frac{1}{3} = \frac{5}{3}$ The point at which normal to be drawn

is (pi/3 , 5/3) ; f(x)=sec x- cot ^2 x

$f ' \left(x\right) = \sec x \tan x - 2 \cot x \cdot \left(- {\csc}^{2} x\right)$ or

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

$f ' \left(\frac{\pi}{4}\right) = \sec \left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{4}\right) + 2 \cot \left(\frac{\pi}{4}\right) \cdot {\csc}^{2} \left(\frac{\pi}{4}\right)$or

$f ' \left(\frac{\pi}{4}\right) = 2 \sqrt{3} + \frac{2}{\sqrt{3}} \cdot \frac{4}{3} = 5.00$ The slope of

tangent at $\frac{\pi}{3}$ is $5.0$ therefore, slope of normal at $\frac{\pi}{3}$

is $\left(- \frac{1}{5}\right)$ The equation of normal to $f \left(x\right)$ at point

$\left(\frac{\pi}{3} , \frac{5}{3}\right)$ is $y - \frac{5}{3} = - \frac{1}{5} \left(x - \frac{\pi}{3}\right)$ or

$15 y - 25 = - 3 x + \pi \mathmr{and} 3 x + 15 y = 25 + \pi$ or

$3 x + 15 y = 28.14$

The equation of normal is $3 x + 15 y = 28.14$ [Ans]

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