Test Index

MHTCET 13 Apr 2026 Shift 2 Paper

© examsnet.com

Question : 114 of 150

Marks: +1, -0

If

A=\left[\begin{array}{cc}\sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta\end{array}\right], \theta \in\left(0, \frac{\pi}{2}\right)

A+\operatorname{adj} A=4 I

\theta=

$\frac{\pi}{12}$
$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$

Solution:

Concept:

For a

2 \times 2

matrix,

\operatorname{adj} A

swaps the diagonal entries and changes the signs of the off-diagonal entries.

Explanation:

Given

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

.

So,

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

.

Now,

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

.

Given

A+\operatorname{adj} A=4I

, we get

2\sec\theta=4

.

Thus,

\sec\theta=2

, so

\cos\theta=\frac{1}{2}

.

Since

\theta\in\left(0,\frac{\pi}{2}\right)

, we have

\theta=\frac{\pi}{3}

.

Answer:

\theta=\frac{\pi}{3}

, which is Option C.

© examsnet.com

Go to Question:

Prev Question

Next Question