Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 118 / 4 / e / Solution

Because the two Chern connections have the same $(0,1)$ part, $a=D_1-D_2$ has type $(1,0)$. The curvature difference formula has only $(2,0)$ and $(1,1)$ parts. Both Chern curvatures have type $(1,1)$, so the total $(2,0)$ part vanishes. The remaining part is obtained from $D_2^{0,1}={\bar{\partial }}_{\mathrm{End}E}$, proving the curvature difference of two Chern connections