The coslice category $C/\bf{C}$ under an object $C \in \bf{C}$ has as objects arrows $f$ in $\bf{C}$ such that $\textbf{dom}(f) = C$ and as arrows, arrows $a: f \to g$ which are arrows $a$ in $ \bf{C}$ such that $af = g$.
So I want to construct this coslice category from the slice category $\textbf{C}/C$ and the $\textbf{op}$ operator (opposite or "dual" categories).
$\textbf{C}/C$ is the same thing as $C/\bf{C}$ except in the definition $\textbf{cod}(f) = C$ and $f = ga$.
So my guess is that $\textbf{C}^{\text{op}}/C = C/\textbf{C}$ since $(\textbf{C}/C)^{\text{op}}$ would first discard some arrows, namely those strictly coming from $C$ (not going to).
But intuitively we need take the $\textbf{op}$ of the whole thing: $(\textbf{C}^{\text{op}}/C)^{\text{op}}$ since otherwise all of our objects would be arrows pointing at $C$ not under $C$.
Therefore that is my final guess. Now what would suffice as a proof that the two are equal: $(\textbf{C}^{\text{op}}/C)^{\text{op}} = C/\bf{C}$?
Objects of $(\mathbf{C}^{\mathrm{op}}/C)^{\mathrm{op}}$ are by definition the morphisms $D\to C$ in $\mathbf{C}^{\mathrm{op}}$, which are by definition the morphisms $C\to D$ in $\mathbf{C}$, that is, the objects of $C/\mathbf{C}$.
A morphism $(f:D\to C)\to (g:E\to C)$ in the same category is by definition a morphism $a:g\to f$ in $\mathbf{C}^{\mathrm{op}}/C$, that is, a commutative triangle $f\circ a=g$ in $\mathbf{C}^{\mathrm{op}}$. Such triangles are by definition the triangles $a\circ f=g$ in $\mathbf{C}$, that is, the morphisms $f\to g$ in $C/\mathbf{C}$.