[proofplan] We translate the assertion into the rank function of the dual matroid. The dual rank formula relates the rank of $C^*$ in $M^*$ to the rank of its complement in $M$, so minimal dependence of $C^*$ in $M^*$ becomes the condition that $E \setminus C^*$ has rank one less than $E$ and that adjoining any missing element raises the rank to full rank. That rank condition is exactly the characterization of hyperplanes as maximal proper flats. [/proofplan]

[step:Record the dual rank formula and the rank criterion for hyperplanes]Let $r_M:2^E \to \mathbb{N}\cup\{0\}$ denote the rank function of $M$, and let $M^*$ be the dual matroid with rank function $r_{M^*}:2^E \to \mathbb{N}\cup\{0\}$. For every subset $A\subset E$, the dual rank formula is \begin{align*} r_{M^*}(A)=|A|-r_M(E)+r_M(E\setminus A). \end{align*} We will use the following rank characterization: a subset $H \subset E$ is a hyperplane of $M$ if and only if \begin{align*} r_M(H)=r_M(E)-1 \end{align*} and, for every $e \in E\setminus H$, \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} Indeed, if $H$ is a maximal proper flat, then $r_M(H)<r_M(E)$. For each $e\in E\setminus H$, maximality of $H$ as a proper flat forces the flat closure of $H\cup\{e\}$ to be $E$, hence $r_M(H\cup\{e\})=r_M(E)$. Since adding one element raises rank by at most one, $r_M(H)=r_M(E)-1$. Conversely, suppose the two displayed rank conditions hold. Since $r_M(H\cup\{e\})>r_M(H)$ for every $e\in E\setminus H$, no element outside $H$ lies in the closure of $H$, so $H$ is a flat. Also $r_M(H)<r_M(E)$, so $H\ne E$. If $F$ is a flat with $H\subsetneq F\subset E$, choose $e\in F\setminus H$. Then \begin{align*} r_M(F)\ge r_M(H\cup\{e\})=r_M(E). \end{align*} Thus $F$ has full rank, so its closure is $E$; since $F$ is flat, $F=E$. Therefore $H$ is a maximal proper flat, hence a hyperplane.[/step]

[guided]The proof will use one bridge between $M$ and its dual $M^*$: the rank of a subset $A\subset E$ in the dual matroid is computed from the rank of the complement $E\setminus A$ in $M$. We denote the rank functions by $r_M$ and $r_{M^*}$, and the formula is \begin{align*} r_{M^*}(A)=|A|-r_M(E)+r_M(E\setminus A). \end{align*} We also isolate the exact rank condition that says a set is a hyperplane. A hyperplane is a maximal proper flat. If $H$ is such a flat, then $r_M(H)<r_M(E)$ because a flat of full rank has closure equal to $E$, and a flat equals its closure. For any $e\in E\setminus H$, maximality says that the only flat properly containing $H$ and containing $e$ is $E$ itself. Equivalently, the closure of $H\cup\{e\}$ is $E$, so \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} Adding one element can increase rank by at most one, so the strict inequality $r_M(H)<r_M(H\cup\{e\})=r_M(E)$ forces \begin{align*} r_M(H)=r_M(E)-1. \end{align*} For the converse direction of this criterion, assume \begin{align*} r_M(H)=r_M(E)-1 \end{align*} and \begin{align*} r_M(H\cup\{e\})=r_M(E) \end{align*} for every $e\in E\setminus H$. The second condition says that adjoining any outside element strictly increases the rank, so no outside element belongs to the closure of $H$. Hence $H$ equals its closure and is a flat. The first condition says $H$ is proper. Finally, if a flat $F$ satisfies $H\subsetneq F\subset E$, then choosing $e\in F\setminus H$ gives \begin{align*} r_M(F)\ge r_M(H\cup\{e\})=r_M(E). \end{align*} Thus $F$ has full rank, so its closure is $E$. Because $F$ is flat, $F=E$. Therefore no proper flat lies strictly between $H$ and $E$, and $H$ is a hyperplane.[/guided]

[step:Show that the complement of a cocircuit satisfies the hyperplane rank criterion]Assume that $C^*$ is a cocircuit of $M$. By definition, $C^*$ is a circuit of $M^*$, so $C^*$ is dependent in $M^*$ and $C^*\setminus\{e\}$ is independent in $M^*$ for every $e\in C^*$. Define \begin{align*} H:=E\setminus C^*. \end{align*} Since a circuit has rank one less than its cardinality, \begin{align*} r_{M^*}(C^*)=|C^*|-1. \end{align*} Using the dual rank formula with $A=C^*$ gives \begin{align*} |C^*|-1=|C^*|-r_M(E)+r_M(H). \end{align*} Cancelling $|C^*|$ yields \begin{align*} r_M(H)=r_M(E)-1. \end{align*} Now fix $e\in C^*$. Since $C^*\setminus\{e\}$ is independent in $M^*$, \begin{align*} r_{M^*}(C^*\setminus\{e\})=|C^*|-1. \end{align*} The complement of $C^*\setminus\{e\}$ in $E$ is $H\cup\{e\}$, so the dual rank formula gives \begin{align*} |C^*|-1=|C^*|-1-r_M(E)+r_M(H\cup\{e\}). \end{align*} Cancelling $|C^*|-1$ gives \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} Since this holds for every $e\in C^*=E\setminus H$, the rank criterion from the previous step shows that $H=E\setminus C^*$ is a hyperplane of $M$.[/step]

[guided]We start from the definition of a cocircuit: $C^*$ is a circuit in the dual matroid $M^*$. Thus $C^*$ is dependent in $M^*$, but deleting any element makes it independent. Define the complement \begin{align*} H:=E\setminus C^*. \end{align*} The goal is to verify the two rank conditions from the previous step for this set $H$. Because $C^*$ is a circuit in $M^*$, its rank is one less than its cardinality: \begin{align*} r_{M^*}(C^*)=|C^*|-1. \end{align*} Now apply the dual rank formula with $A=C^*$. Since $E\setminus C^*=H$, we get \begin{align*} |C^*|-1=|C^*|-r_M(E)+r_M(H). \end{align*} Cancelling the common term $|C^*|$ gives \begin{align*} r_M(H)=r_M(E)-1. \end{align*} This is the first hyperplane rank condition. It remains to show that adjoining any element outside $H$ raises the rank to full rank. The elements outside $H$ are precisely the elements of $C^*$. Fix $e\in C^*$. Since $C^*$ is a circuit in $M^*$, the set $C^*\setminus\{e\}$ is independent in $M^*$. Therefore its rank equals its cardinality: \begin{align*} r_{M^*}(C^*\setminus\{e\})=|C^*|-1. \end{align*} The complement of $C^*\setminus\{e\}$ in $E$ is $H\cup\{e\}$. Applying the dual rank formula to $A=C^*\setminus\{e\}$ gives \begin{align*} |C^*|-1=|C^*|-1-r_M(E)+r_M(H\cup\{e\}). \end{align*} Cancelling $|C^*|-1$ yields \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} Since $e\in C^*$ was arbitrary and $C^*=E\setminus H$, every element outside $H$ raises the rank to $r_M(E)$. The rank criterion from the previous step therefore implies that $H=E\setminus C^*$ is a hyperplane of $M$.[/guided]

[step:Show that the complement of a hyperplane is a cocircuit]Assume that $H:=E\setminus C^*$ is a hyperplane of $M$. Since a hyperplane is a proper subset of $E$, we have $H\ne E$, and therefore $C^*=E\setminus H$ is nonempty. By the rank criterion for hyperplanes, \begin{align*} r_M(H)=r_M(E)-1 \end{align*} and, for every $e\in C^*$, \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} Applying the dual rank formula to $A=C^*$ gives \begin{align*} r_{M^*}(C^*)=|C^*|-r_M(E)+r_M(H)=|C^*|-1. \end{align*} Thus $C^*$ is dependent in $M^*$. Now fix $e\in C^*$. Since $E\setminus(C^*\setminus\{e\})=H\cup\{e\}$, the dual rank formula gives \begin{align*} r_{M^*}(C^*\setminus\{e\})=|C^*|-1-r_M(E)+r_M(H\cup\{e\})=|C^*|-1. \end{align*} Hence $C^*\setminus\{e\}$ is independent in $M^*$. Every proper subset $S\subsetneq C^*$ is contained in $C^*\setminus\{e\}$ for some $e\in C^*\setminus S$, and independence is hereditary in a matroid. Therefore every proper subset of $C^*$ is independent in $M^*$. Since $C^*$ itself is dependent in $M^*$, it is a circuit of $M^*$. By definition, $C^*$ is a cocircuit of $M$.[/step]

[guided]Now assume that the complement $H:=E\setminus C^*$ is a hyperplane of $M$. A hyperplane is proper, so $H\ne E$ and hence $C^*=E\setminus H$ is nonempty. By the rank criterion proved in the first step, this means \begin{align*} r_M(H)=r_M(E)-1 \end{align*} and, for every $e\in C^*$, \begin{align*} r_M(H\cup\{e\})=r_M(E). \end{align*} We must prove that $C^*$ is a circuit of the dual matroid $M^*$, meaning that $C^*$ is dependent in $M^*$ and every proper subset of $C^*$ is independent in $M^*$. First apply the dual rank formula to $A=C^*$. Since $E\setminus C^*=H$, the hyperplane rank condition gives \begin{align*} r_{M^*}(C^*)=|C^*|-r_M(E)+r_M(H)=|C^*|-1. \end{align*} A set is independent exactly when its rank equals its cardinality. Here the rank of $C^*$ in $M^*$ is strictly smaller than $|C^*|$, so $C^*$ is dependent in $M^*$. Next fix $e\in C^*$. The complement of $C^*\setminus\{e\}$ is $H\cup\{e\}$. Applying the dual rank formula again, now to $A=C^*\setminus\{e\}$, and using the second hyperplane rank condition gives \begin{align*} r_{M^*}(C^*\setminus\{e\})=|C^*|-1-r_M(E)+r_M(H\cup\{e\})=|C^*|-1. \end{align*} This equals the cardinality of $C^*\setminus\{e\}$, so $C^*\setminus\{e\}$ is independent in $M^*$. Finally let $S\subsetneq C^*$ be any proper subset. Choose $e\in C^*\setminus S$. Then $S\subset C^*\setminus\{e\}$, and independence is hereditary in a matroid, so $S$ is independent in $M^*$. Thus every proper subset of $C^*$ is independent while $C^*$ itself is dependent. Hence $C^*$ is a circuit of $M^*$, and by definition $C^*$ is a cocircuit of $M$.[/guided]

[step:Conclude the equivalence] We have shown that if $C^*$ is a cocircuit of $M$, then $E\setminus C^*$ is a hyperplane of $M$, and that if $E\setminus C^*$ is a hyperplane of $M$, then $C^*$ is a cocircuit of $M$. This proves the desired equivalence. [/step]