Question

If $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$ and $\underline{PQ=I}$, then

$(cosec^2 \theta) Q$ is given by:

(where $I$ is an identity matrix of $2 \times 2$ order)

• A. P($\theta$)
• B. P(-$\theta$)
• C. P(2$\theta$)
• D. I

Answer 1

Answer: (B)

P(-$\theta$)

Answer 2

Given

$$P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}.$$

Since $PQ = I$, we have

$$Q = P(\theta)^{-1} = \frac{1}{\det P(\theta)} \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}.$$

Calculate the determinant:

$$\det P(\theta) = 1\cdot1 - (\cot \theta)(-\cot \theta) = 1 + \cot^2\theta = \csc^2 \theta.$$

Thus,

$$Q = \frac{1}{\csc^2\theta}\begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix} = \sin^2\theta \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}.$$

Multiplying both sides by $\csc^2 \theta$ we get:

$$(\csc^2 \theta) Q = \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}.$$

Now, note that:

$$P(-\theta) = \begin{bmatrix} 1 & \cot(-\theta) \\ -\cot(-\theta) & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}.$$

Thus,

$$(\csc^2 \theta) Q = P(-\theta).$$