Physics..

That's a no-no. Only reason why it doesn't matter this time is because they completely cancel each other out.

 Alright, 2nd attempt, or so help me jebus!

Code:

[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 - 2|A||B|cos(180-θ)
[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 + 2|A||B|cos(θ)
[2|A|^2 - 2|A|^2*cos(θ)]n^2 = 2|A|^2 + 2|A|^2cos(θ)
[1 - cos(θ)]n^2 = 1 + cos(θ)
n^2 = 1 + cos(θ) + cos(θ)*n^2
n^2 = 1 + (1 + n^2)*cos(θ)
(n^2 - 1) / (n^2 + 1) = cos(θ)

tbh, I'm not sure how the other guy managed to get θ = 2*arctan(1/n) @_@ I like his answer though.

Nicely done. Only thing left to do is to use half angle formula for cosine, and you'll get θ = 2*arctan(1/n)

 Spoiler

cos(θ = [1-(tan(θ/2))^2]/[1+(tan(θ/2))^2]

Edit: I completely missed the part where you defined your θ as the angle between vector A and -B.

The angle between vector A and B would be equal to [180 - 2*arctan(1/n)] or simply 2*arctan(n)