Reaction of trans-2-phenyl-1-bromocyclopentane with alcoholic $\text{KOH}$ produces:

This is an $\text{E}2$ elimination reaction. $\text{E}2$ reactions require an anti-periplanar orientation of the Leaving Group (Br) and the beta-hydrogen.In trans-2-phenyl-1-bromocyclopentane, the Br and the Phenyl group are on opposite sides. This means the Hydrogen on C-2 (where Phenyl is) is cis to the Bromine.Therefore, the $\text{E}2$ elimination cannot happen using the hydrogen on C-2. It must use the hydrogen on C-5.Elimination between C-1 and C-5 gives 3-phenylcyclopentene.