If P(theta) = begin{bmatrix} 1 & cot theta - | vedclass

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(B) Given $P(	heta) = \begin{bmatrix} 1 & \cot 	heta \ -\cot 	heta & 1 \end{bmatrix}$.First,we find the determinant of $P(	heta)$:$|P(	heta)| =

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$P(-	heta)$

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(B) Given $P(	heta) = \begin{bmatrix} 1 & \cot 	heta \ -\cot 	heta & 1 \end{bmatrix}$.First,we find the determinant of $P(	heta)$:$|P(	heta)| = (1)(1) - (\cot 	heta)(-\cot 	heta) = 1 + \cot^2 	heta = \csc^2 	heta$.Since $PQ = I$,$Q$ is the inverse of $P$,i.e.,$Q = P^{-1} = \frac{1}{|P(	heta)|} 	ext{adj}(P(	heta))$.The adjoint of $P(	heta)$ is $\begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Thus,$Q = \frac{1}{\csc^2 	heta} \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.We need to find $(\csc^2 	heta)Q$:$(\csc^2 	heta)Q = \csc^2 	heta \cdot \frac{1}{\csc^2 	heta} \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Now,observe $P(-	heta) = \begin{bmatrix} 1 & \cot(-	heta) \ -\cot(-	heta) & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Therefore,$(\csc^2 	heta)Q = P(-	heta)$.

If $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$ and $PQ = I$,then find $(\csc^2 \theta)Q$,where $I$ is an identity matrix of order $2 \times 2$.

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