What is the equation of the line that is normal to $f (x) = csc x + tan x - cot x$ $f(x)=cscx+tanx-cotx$ at $x = - \frac{π}{3}$ $x=-pi/3$ ?

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Answer 1 · 2018-02-23T11:25:01

$y = - \frac{3 x}{14} - 2.53$ $y=-(3x)/14-2.53$

$"Tangent":d/dx[f(x)]=f'(x)$

$"Normal":-1/(f'(x))=-1/(d/dx[cscx+tanx-cotx])=-1/(d/dx[cscx]+d/dx[tanx]-d/dx[cotx])=-1/(-cscxcotx+sec^2x+csc^2x)$

$-1/(f'(-pi/3))=-1/(-csc(-pi/3)cot(-pi/3)+sec^2(-pi/3)+csc^2(-pi/3))=-1/(14/3)=-3/14$

$y=mx+c$

$f(a)=ma+c$

$csc(-pi/3)+tan(-pi/3)-cot(-pi/3)=-pi/3(-3/14)+c$

$c=csc(-pi/3)+tan(-pi/3)-cot(-pi/3)+pi/3(-3/14)$

$c=-2.53$

$y=-(3x)/14-2.53$

What is the equation of the line that is normal to f(x)=cscx+tanx-cotx f(x)=cscx+tanx−cotxf(x)=cscx+tanx-cotx at x=-pi/3x=−π3 x=-pi/3?