On page 21 of Continuum mechanics by C. S. Jog the author states that
$${\bf cof \: T}^T = ({\bf cof \: T)}^T$$
I've found a proof here on mathSE that's based on determinants but this relationship is not introduced until page 22 of the text, which suggests to me that there is another proof that is not based on determinants.
The approach I tried is as follows:
$$ {\bf cof \: T}^T ({\bf u} \; \times {\bf v}) = {\bf T^T u} \; \times {\bf T^T v} $$
whereas
$$ ({\bf cof \: T})^T ({\bf u} \; \times {\bf v}) = ({\bf T\, u} \; \times {\bf T \, v})^T$$
Unfortunately this does not lead anywhere. Besides the dimension of the first relation appears to be of order $n \times 1$, whereas the second relation has dimension $1 \times n$, if $\bf T$ is of order $n$.
On the same page author states the relation
$$({\bf cof \: T})^T = I_2 {\bf \, I} - (tr {\bf \, T})\, {\bf T} + {\bf T}^2$$
Setting ${\bf T}^T$ for ${\bf T}$ we get
$$ \begin{equation}\begin{aligned} ({\bf cof \: T}^T)^T & = I_2 {\bf \, I} - (tr {\bf \, T}^T)\, {\bf T}^T + ({\bf T}^T)^2 \\ & = I_2 {\bf \, I} - (tr {\bf \, T})\, {\bf T}^T + ({\bf T}^2)^T \end{aligned}\end{equation} $$
because that $({\bf T}^T)^2 = ({\bf T}^2)^T$. Taking the transpose of both sides results in the orginal expression, showing that ${\bf cof \; T}^T$ = $({\bf cof \; T)}^T$
This proof assumes that $I_2$ is invariant under the change ${\bf T}^T$ to ${\bf T}$.