RBC model and indivisible labour
The only reason I can see right now could possibly be pedagogical. While that term does indeed just equal zero, there could be generalizations where the agent works a high amount with probability $\tau$ and a low, but positive, amount with probability $1-\tau$. In this case, the term would still go through. There are also plenty of functional forms that would result in that term not being zero (taking away the -1, for example, which is often done in applied work). At any rate, this is an idiosyncratic enough case that the professor might have thought it best to show the process of rearranging terms for future reference and to give help with other scenarios. On the other hand, he may just not have noticed it. :)