1. We first identify the Werner coordination formulas for each compound.

For cobalt(III), the coordination number is typically $6$.

2. Analyze each complex.

(a) $CoCl_3\cdot 4NH_3$

This corresponds to the complex: $[Co(NH_3)_4Cl_2]Cl$ Here, the coordination sphere has $4NH_3$ and $2Cl^-$, i.e. an octahedral complex of type: $[MA_4B_2]$ Such octahedral complexes show cis-trans isomerism.

So, (a) shows cis-trans isomerism.

(b) $CoCl_3\cdot 5NH_3$

This corresponds to: $[Co(NH_3)_5Cl]Cl_2$ This is an octahedral complex of type: $[MA_5B]$ Such complexes do not show cis-trans isomerism because there is only one ligand of type $B$.

So, (b) does not show cis-trans isomerism.

(c) $CoCl_3\cdot 6NH_3$

This corresponds to: $[Co(NH_3)_6]Cl_3$ This is of type: $[MA_6]$ All ligands are identical, so no geometrical isomerism is possible.

So, (c) does not show cis-trans isomerism.

(d) $CoCl(NO_3)_2\cdot 5NH_3$

Total anions outside/inside must satisfy coordination number $6$ for $Co(III)$. This gives the complex as: $[Co(NH_3)_5Cl](NO_3)_2$ or $[Co(NH_3)_5(NO_3)]Cl(NO_3)$ But in either case, the coordination sphere is of type $[MA_5B]$.

An octahedral $[MA_5B]$ complex does not show cis-trans isomerism.

So, (d) does not show cis-trans isomerism.

3. Count the number of complexes showing cis-trans form.

Only (a) shows cis-trans isomerism.

Therefore, the required number is: $\boxed{1}$