The teacher hasn't uploaded a solution for this question yet.
Given $A=\begin{bmatrix}1&cot~x\\ -cot~x&1\end{bmatrix}$, the transpose $A'$ is obtained by interchanging rows and columns: $$A' = \begin{bmatrix}1&-cot~x\\ cot~x&1\end{bmatrix}$$
The determinant of $A$ is: $$|A| = (1)(1) - (cot~x)(-cot~x) = 1 + cot^2x = cosec^2x$$
The inverse of $A$ is given by: $$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{cosec^2x} \begin{bmatrix}1&-cot~x\\ cot~x&1\end{bmatrix} = sin^2x \begin{bmatrix}1&-cot~x\\ cot~x&1\end{bmatrix} = \begin{bmatrix}sin^2x&-sin^2x~cot~x\\ sin^2x~cot~x&sin^2x\end{bmatrix} = \begin{bmatrix}sin^2x&-sinx~cosx\\ sinx~cosx&sin^2x\end{bmatrix}$$
Now, we compute $A'A^{-1}$: $$A'A^{-1} = \begin{bmatrix}1&-cot~x\\ cot~x&1\end{bmatrix} \begin{bmatrix}sin^2x&-sinx~cosx\\ sinx~cosx&sin^2x\end{bmatrix} = \begin{bmatrix}sin^2x - cotx~sinx~cosx & -sinx~cosx - cotx~sin^2x\\ cotx~sin^2x + sinx~cosx & -cotx~sinx~cosx + sin^2x\end{bmatrix} = \begin{bmatrix}sin^2x - cos^2x & -2sinx~cosx\\ 2sinx~cosx & sin^2x - cos^2x\end{bmatrix} = \begin{bmatrix}-cos~2x&-sin~2x\\ sin~2x&-cos~2x\end{bmatrix}$$
Final Answer: $A^{\prime}A^{-1}=\begin{bmatrix}-cos~2x&-sin~2x\\ sin~2x&-cos~2x\end{bmatrix}$
AI generated content. Review strictly for academic accuracy.