Suppose $X$ is a DM stack, and let $E^\bullet$ be a perfect obstruction theory of $X$ such that the $E^{-1}$ term admits a trivial quotient/sub-bundle. Is it true that the virtual fundamental class $[X, E^\bullet]$ is zero?

If $X$ is smooth, then this is true: In such a case, the virtual fundamental class is the top Chern class of the vector bundle $E^{-1}$, which is zero due to the trivial quotient/sub-bundle and the exact sequence $$0 \to \mathcal{O} \to E^{-1} \to coker \to 0$$ together with the multiplicative nature of $c_{top}$. Intuitively, you can use the trivial factor to "move" a section of this bundle away from the zero section.

If $X$ is not smooth, then this argument doesn't work and we must use the intrinsic normal cone of Behrend and Fantechi to compute the virtual fundamental class. Instead of bundles, we obtain a cone $C(E^\bullet)$ contained in $(E^{-1})^\vee$, which we intersect with the zero section of $(E^{-1})^\vee$. In comparison with the smooth case, it seems like we should somehow be able to move the cone out of the zero section using the trivial portion of the bundle, but I don't see how to do this.

Is there an easy argument showing that this class is zero? Are there extra conditions required?

**Edit**: The method which I have tried to no avail is to use Proposition 5.10 of Behrend-Fantechi. It states loosely that, given two obstruction theories $F$ and $F'$, and certain compatibility data between them, that
$$
v^![X,F] = [X,F']
$$
However, I was not able to find a clear way to have this yield that my desired virtual class is zero.