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

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Answer 1 · 2018-02-23T11:25:01

#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#