Let $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, then $(A^{-1})^T$ equals

$A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

For a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is:

$\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

The determinant of $A$:

$\text{det}(A) = \cos\theta \cdot \cos\theta - (-\sin\theta) \cdot \sin\theta$

$= \cos^2\theta + \sin^2\theta$

$= 1$

---

The inverse of $A$:

$A^{-1} = \frac{1}{1}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$

---

The transpose of $A^{-1}$ flips the matrix along its main diagonal:

$(A^{-1})^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

This equals the original matrix $A$.

Therefore, $(A^{-1})^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$