Step-by-step solution:

1. Identify the Reactants and Reaction:

   • The reactant is $trans$-2-butene, an alkene with the chemical formula $CH_3CH=CHCH_3$. The trans configuration means the two methyl ($CH_3$) groups are on opposite sides of the double bond.
   • The reaction is bromination, which is the addition of a bromine molecule ($Br_2$) across the double bond.

2. Determine the Reaction Mechanism and Stereochemistry:

   • The addition of halogens like $Br_2$ to an alkene is an electrophilic addition reaction.
   • The mechanism involves the formation of a cyclic bromonium ion intermediate. The subsequent attack by the bromide ion ($Br^−$) occurs from the side opposite to the cyclic intermediate. This is known as anti-addition.

3. Analyze the Stereochemical Outcome:

   • We are starting with a trans alkene and performing an anti addition. There is a general rule for the stereochemical outcome of such reactions on symmetrical alkenes:
     • cis + syn addition → meso
     • cis + anti addition → racemic (pair of enantiomers)
     • trans + syn addition → racemic (pair of enantiomers)
     • trans + anti addition → meso
   • Following this rule, the bromination of $trans$-2-butene (a trans alkene) via anti addition should yield a meso compound.

4. Illustrate the Reaction Product:

   • The product of the reaction is 2,3-dibromobutane ($CH_3-CH(Br)-CH(Br)-CH_3$). This molecule has two chiral centers at carbon-2 and carbon-3.

   • Let's visualize the reaction using Fischer projections to confirm the formation of a meso compound.

   • First, draw the structure of $trans$-2-butene in a way that is suitable for showing the addition. The main carbon chain is vertical. $\begin{array}{c} CH_3 \\ | \\ H-C \\ || \\ C-H \\ | \\ CH_3 \end{array}$

   • In an anti-addition, the two bromine atoms add from opposite faces of the double bond. In a Fischer projection, this means they will appear on opposite sides (one on the left, one on the right). $\begin{array}{c} CH_3 \\ | \\ Br - C - H \\ ---|--- \\ H - C - Br \\ | \\ CH_3 \end{array}$

5. Identify the Product as a Meso Compound:

   • The resulting structure is a Fischer projection of 2,3-dibromobutane.
   • Let's analyze this structure for elements of symmetry. A meso compound is an achiral compound that has chiral centers. It is achiral because it possesses an internal plane of symmetry or a center of inversion in one of its conformations.
   • The Fischer projection shown for the product has a center of inversion. If you trace a line from any group through the midpoint of the C2-C3 bond, you find an identical group at an equal distance on the other side. For example, the top $CH_3$ and bottom $CH_3$, the left $Br$ and right $Br$, the right $H$ and left $H$.
   • A molecule with a center of inversion is achiral. This specific stereoisomer is called meso-2,3-dibromobutane.
   • Alternatively, we can assign the R/S configuration to the chiral centers. The top chiral center (C2) is (S), and the bottom chiral center (C3) is (R). The molecule is (2S, 3R)-2,3-dibromobutane. Its mirror image would be (2R, 3S)-2,3-dibromobutane. For a symmetrical molecule like this, the (2S, 3R) and (2R, 3S) forms are identical and represent the same single meso compound.

6. Conclusion:

   • The bromination of $trans$-2-butene proceeds via anti-addition to form a single product, the meso-2,3-dibromobutane.
   • A meso compound is a single stereoisomer.
   • Therefore, the number of stereoisomers obtained is 1.

Final Answer Check:

• The question asks for the number of stereoisomers obtained.
• Our analysis shows that only one product, a meso compound, is formed.
• The number of stereoisomers is 1.
• This corresponds to option A.