What is the slope of the line normal to the tangent line of $f (x) = csc x - {sec}^{2} x + 1$ $f(x) = cscx-sec^2x +1$ at $x = \frac{17 π}{12}$ $x= (17pi)/12$ ?

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What is the slope of the line normal to the tangent line of $f (x) = csc x - {sec}^{2} x + 1$ $f(x) = cscx-sec^2x +1$ at $x = \frac{17 π}{12}$ $x= (17pi)/12$ ?

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Answer 1 · 2018-06-02T13:35:50

$y = - 1 (6 \cdot \sqrt{6}) \cdot x + \frac{1}{432} \cdot (432 + 864 \cdot \sqrt{2} + 17 \cdot \sqrt{6} \cdot π)$ $y=-1(6*sqrt(6))*x+1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)$

Explanation:

We get
$f'(x)=-cot(x)*csc(x)-sec(x)*tan(x)$
so
$f'(17*pi/12)=6*sqrt(6)$
and the slope of the normal line is given by
$m_N=-1/(6*sqrt(6))$
and
$f(17/12*pi)=1+2sqrt(2)$
From the equation

$1+2sqrt(2)=-1/(6*sqrt(6))*17/12*pi+n$
we get
$n=1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)$

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What is the slope of the line normal to the tangent line of $f (x) = csc x - {sec}^{2} x + 1$ $f(x) = cscx-sec^2x +1$ at $x = \frac{17 π}{12}$ $x= (17pi)/12$ ?

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What is the slope of the line normal to the tangent line of f(x) = cscx-sec^2x +1 f(x)=cscx−sec2x+1f(x) = cscx-sec^2x +1 at x= (17pi)/12 x=17π12 x= (17pi)/12 ?

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What is the slope of the line normal to the tangent line of $f (x) = csc x - {sec}^{2} x + 1$ $f(x) = cscx-sec^2x +1$ at $x = \frac{17 π}{12}$ $x= (17pi)/12$ ?