Why does dichloridobisethylenediaminecobalt(III) have only three optical isomers?
In simplistic terms, we can see $\ce{[Co(en)2Cl2]^+}$ as an $\ce{[MA4B2]}$ pseudooctahedral complex. We can initially consider the $\ce{A\bond{-&gt;}M}$ and $\ce{B\bond{-&gt;}M}$ bonds identical. Trying this out on a model octahedron or molecular modelling kit will show you that only two possible different configurations exist: cis and trans.

Here, however, we have one of the ligand types actually being a briding ligand. Because the bridge is too short to allow for a trans configuration, the two coordination sites of these bridging ligands will always be cis to each other.

For the complex in which the two chlorido ligands are trans to each other, only one possible arrangement of the $\ce{en}$ ligands is possible. At first approximation, the entire complex should have $C_\mathrm{2v}$ symmetry if not more: a $C_2$ axis with two perpendicular $\sigma_\mathrm{v}$ planes; one bisecting the $\ce{en}$ ligands through the $\ce{C-C}$ bond, the other being orthogonal and mapping each $\ce{en}$ onto the other. Due to the presence of planes of symmetry, this complex cannot be chiral.

In the complex in which the two chlorido ligands are cis to each other, one could think of up to three potential arrangements for the bridges; however, as I pointed out, it is not possible to have the two coordinating atoms of one $\ce{en}$ ligand be trans to each other, reducing the total down to two. These two possibilities do not differ in whether one of the chlorides is in an ‘axial’ position (remember, we are assuming equal bond lengths initially so there is no defined ‘axial’ position). Rather, in one case you have a leftwards-turning screw and in the other case a rightwards-turning screw if you assume the $\ce{en}$ ligand to be a plane and wanted to screw the entire thing into your paper.

Neither of these two structures have a plane of symmetry (or higher symmetry) and both are therefore optically active. Careful analysis will reveal, that they are in fact mirror images and thus enantiomers of one another. They are called the &Delta; and &Lambda; isomers, respectively. All three isomers are attached in the image below.


Figure 1: isomers of $\ce{[Co(en)2Cl2]^+}$. From left to right: trans, cis-&Lambda;, cis-&Delta;.

Thus, there are three isomers of the compound, two of which are optical isomers of each other (the third is a geometrical isomer).

Finally on the topic of whether axial bonds are actually longer, i.e. whether there is a Jahn-Teller type distortion or not. The answer is no. $\ce{Co^{III}}$ is a $\mathrm{d^6}$ complex. It could be assumed to be high or low spin. In case it accidentally ends up to be low spin, the $\mathrm{t_{2g}}$ orbitals are fully populated while the $\mathrm{e_g}$ orbitals are entirely unpopulated; there is no need for Jahn-Teller distortion.

In the potentially more likely high-spin case, one of the $\mathrm{t_{2g}}$ orbitals is fully populated while all other orbitals are half-populated. We might expect a Jahn-Teller distortion to remove the degeneracy of the $\mathrm{t_{2g}}$ orbitals. However, this distortion would be minor at best: the $\mathrm{t_{2g}}$ orbitals do not interact well with the ligands since they are oriented to point in-between the ligand orbitals. (On a group theory basis, the $\mathrm{t_{2g}}$ orbitals are nonbonding.) Therefore, any symmetry reducing effect will be very minor at best and not lead to a noticeable distortion. Thus, it is reasonable to assume near-identical bond lengths and no principal (‘axial’) direction.