1. The problem statement, all variables and given/known data arccot x = (π/2) - arctan x arccot x =/= π arccot x =/= 0 2. Relevant 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) 3. 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.