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## Homework Statement

arccot x = (π/2) - arctan x

arccot x =/= π

arccot x =/= 0

## Homework Equations

arccot x = 1/arctan x (if x > 0)

arccot x = 1/arctan x + π (if x < 0)

arccot x = π/2 (if x = 0)

## The Attempt at a Solution

π and 0 are the horizontal asymptotes, the values for which y (sine) cannot be.

arccot x = (π/2) - arctan x

arccot x =/= π

π =/= (π/2) - arctan x

(π/2) =/= -arctan x

-(π/2) =/= arctan x

arccot x = (π/2) - arctan x

arccot x =/= 0

0 =/= (π/2) - arctan x

-(π/2) =/= -arctan x

(π/2) =/= arctan x

Therefore, arctan x cannot be equal to negative or positive π/2, because if it were, then adding it to the original equation would produce π or 0, respectively. This cannot work because these are the values that it

*cannot*be. As I write this, I'm thinking that I'm only

*establishing*the fact that 0 and π are the asymptotes. But how would I

*prove*that they're the asymptotes? If it's beyond the scope of precalculus, please say so.

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