Question

If \( A=\begin{bmatrix}\sec\theta & -\tan\theta-\tan\theta & \sec\theta\end{bmatrix} \) and \( A+\operatorname{adj}A=4I \), then find the value of \( \theta \).

• (A) \( \frac{\pi}{6} \)
• (B) \( \frac{\pi}{4} \)
• (C) \( 0 \)
• (D) \( \frac{\pi}{3} \)

Hint:

For symmetric matrices with equal diagonal entries, adding the matrix and its adjoint often simplifies the off-diagonal terms automatically.

Answer 1

The Correct Option is D

To solve the given problem, let's start by analyzing the provided information. We have the matrix \( A \) given by: 

\( A = \begin{bmatrix} \sec\theta & -\tan\theta \\ -\tan\theta & \sec\theta \end{bmatrix} \)

and it is given that:

1. The relation \( A + \operatorname{adj}A = 4I \) holds.

First, we find the adjugate of matrix \( A \), denoted as \( \operatorname{adj}A \). For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjugate is calculated as:

\( \operatorname{adj} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

Thus, for the given matrix \( A \), we have:

\( \operatorname{adj}A = \begin{bmatrix} \sec\theta & \tan\theta \\ \tan\theta & \sec\theta \end{bmatrix} \)

Now, using the condition \( A + \operatorname{adj}A = 4I \), we simply add the entries of \( A \) and \( \operatorname{adj}A \) as follows:

\( A + \operatorname{adj}A = \begin{bmatrix} \sec\theta + \sec\theta & -\tan\theta + \tan\theta \\ -\tan\theta + \tan\theta & \sec\theta + \sec\theta \end{bmatrix} = \begin{bmatrix} 2\sec\theta & 0 \\ 0 & 2\sec\theta \end{bmatrix} \)

Equating to \( 4I = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \), we get:

• \( 2\sec\theta = 4 \)

From this, solve for \( \sec\theta \):

1. \( \sec\theta = 2 \)

Recall that \( \sec\theta = \frac{1}{\cos\theta} \), so:

1. \( \cos\theta = \frac{1}{2} \)

The angle \( \theta \) for which \( \cos\theta = \frac{1}{2} \) is \( \theta = \frac{\pi}{3} \).

Thus, the correct option is:

• (D) \(\frac{\pi}{3}\)