What is the slope of the line normal to the tangent line of $f (x) = {sec}^{2} x - {sin}^{2} x$ $f(x) = sec^2x-sin^2x$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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What is the slope of the line normal to the tangent line of $f (x) = {sec}^{2} x - {sin}^{2} x$ $f(x) = sec^2x-sin^2x$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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Answer 1 · 2018-06-23T09:54:00

$y = (- \frac{74}{11} - 128 \frac{\sqrt{3}}{33}) x + \frac{15}{2} - \frac{15}{4} \sqrt{3} + \frac{37}{66} \cdot π + \frac{32}{99} \cdot π \cdot \sqrt{3}$ $y=(-74/11-128sqrt(3)/33)x+15/2-15/4sqrt(3)+37/66*pi+32/99*pi*sqrt(3)$

Explanation:

$f'(x)=2sec^2(x)tan(x)-2sin(x)cos(x)$

$f'(pi/12)=111/2-32sqrt(3)$
so the slope of the normal line is given by

$-1/(111/2-32sqrt(3))=-74/11-128sqrt(3)/33$
$f(pi/12)=15/8*(sqrt(3)-1)^2$

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What is the slope of the line normal to the tangent line of $f (x) = {sec}^{2} x - {sin}^{2} x$ $f(x) = sec^2x-sin^2x$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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What is the slope of the line normal to the tangent line of f(x) = sec^2x-sin^2x f(x)=sec2x−sin2xf(x) = sec^2x-sin^2x at x= (pi)/12 x=π12 x= (pi)/12 ?

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Explanation:

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What is the slope of the line normal to the tangent line of $f (x) = {sec}^{2} x - {sin}^{2} x$ $f(x) = sec^2x-sin^2x$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?