Mathematically, Cosignϴ x C = B
Cosign ϴ x C (C being the hypoteneuse) this is equal to the direct line of sight distance to target = B (in a right angled triangle). B = the actual horizontal distance to the target. This distance B is the distance that the bullet in flight is effected by gravity not distance C.
The distance C is always by mathematical principal (Pythagoras theorem) a shorter distance than B. Therefore the (measurable) effect of gravity on the bullet is only imparted for distance B.
The rest of your statement is just clouding the issue, the bullet still travels the 1000 mtrs to the target, correct. So the flight time is still your theoretical 2secs, the fact is that, the bullet is only being effected by gravity for 0.866% of the flight, which due to the inclined or declined angle gives a ballistic advantage, the direct result of which is the need to allow for less drop for the given distance.
It doesn't just appear to drop less, it actually does drop less. Your first line in the question relates to the rule used to determine the horizontal distance to target
The solution posted by Jack is quite right, a simple way to determine the distance to target that should be used with a shooters own dope or a ballistics solver to determine the hold for the shot.
The OPs original intent I am sure was to enlighten new shooters to one way to engage targets at inclined angles.