Question

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

Answer 1

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

Explanation:

`"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`