Step 1: Concept
The asymptotes of a hyperbola are $y = \pm(b/a)x$.
Step 2: Analysis
The angle $\alpha$ an asymptote makes with the transverse axis satisfies $\tan(\theta/2) = b/a$.
Step 3: Evaluation
Using the identity $\sec^2(\theta/2) = 1 + \tan^2(\theta/2) = 1 + b^2/a^2 = (a^2+b^2)/a^2$.
Step 4: Conclusion
Since $e^2 = (a^2+b^2)/a^2$, we have $\sec^2(\theta/2) = e^2 \Rightarrow \cos(\theta/2) = 1/e$.
Final Answer: (a)