Question

If \(A = [\begin{array}{cc}secθ & -tanθ \\ -tanθ & secθ\end{array}]\), \(θ\in (0,\frac{π}{2})\) such that \(A+adj\,A = 4I\), then \(θ =\)

• A. \(\frac{π}{12}\)
• B. \(\frac{π}{6}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{4}\)

Hint:

For a 2 by 2 matrix, \(\text{adj}A\) swaps the diagonal and changes signs of the off-diagonal entries.

Answer 1

The Correct Option is C

Step 1: Plan:
The off-diagonal entries cancel in the sum, which makes this quick.

Step 2: Steps:
Cofactors of a 2 by 2 matrix: swap the diagonal, flip the signs of the other two entries. The $-\tan\theta$ entries become $+\tan\theta$, so adding the matrices cancels them.
The diagonal gives $2\sec\theta$ in each place. Equal to $4$ means $\sec\theta = 2$, and $\theta = 60^{\circ} = \frac{\pi}{3}$.
Check the other options: $\sec\frac{\pi}{12}\approx 1.035$, $\sec\frac{\pi}{6}\approx 1.155$, $\sec\frac{\pi}{4} \approx 1.414$, none is $2$.

Final Answer:
The value is $\theta = \frac{\pi}{3}$, option (C). \[ \boxed{\frac{\pi}{3}} \]