Q65 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
If $A = \begin{bmatrix} 1 & -\tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ -\tan\dfrac{\theta}{2} & 1 \end{bmatrix}$ then $A^{-1}B$ is equal to
- Correct answerA$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
- B$\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
- C$\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}$
- D$\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
Solution
Let $t = \tan\dfrac{\theta}{2}$.
The given matrices can be written as:
$A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
The determinant of $A$ is $|A| = 1(1) - (-t)(t) = 1 + t^2$.
The adjugate of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$.
Therefore, the inverse of $A$ is:
$A^{-1} = \dfrac{1}{|A|} \text{adj}(A) = \dfrac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
Now, multiply $A^{-1}$ and $B$:
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1(1) + t(-t) & 1(t) + t(1) \\ -t(1) + 1(-t) & -t(t) + 1(1) \end{bmatrix}$
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1 - t^2 & 2t \\ -2t & 1 - t^2 \end{bmatrix}$
$A^{-1}B = \begin{bmatrix} \dfrac{1-t^2}{1+t^2} & \dfrac{2t}{1+t^2} \\ \dfrac{-2t}{1+t^2} & \dfrac{1-t^2}{1+t^2} \end{bmatrix}$
Using the half-angle trigonometric identities $\cos\theta = \dfrac{1 - \tan^2\dfrac{\theta}{2}}{1 + \tan^2\dfrac{\theta}{2}}$ and $\sin\theta = \dfrac{2\tan\dfrac{\theta}{2}}{1 + \tan^2\dfrac{\theta}{2}}$, we get:
$A^{-1}B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
Answer: $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
The given matrices can be written as:
$A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
The determinant of $A$ is $|A| = 1(1) - (-t)(t) = 1 + t^2$.
The adjugate of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$.
Therefore, the inverse of $A$ is:
$A^{-1} = \dfrac{1}{|A|} \text{adj}(A) = \dfrac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
Now, multiply $A^{-1}$ and $B$:
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1(1) + t(-t) & 1(t) + t(1) \\ -t(1) + 1(-t) & -t(t) + 1(1) \end{bmatrix}$
$A^{-1}B = \dfrac{1}{1+t^2} \begin{bmatrix} 1 - t^2 & 2t \\ -2t & 1 - t^2 \end{bmatrix}$
$A^{-1}B = \begin{bmatrix} \dfrac{1-t^2}{1+t^2} & \dfrac{2t}{1+t^2} \\ \dfrac{-2t}{1+t^2} & \dfrac{1-t^2}{1+t^2} \end{bmatrix}$
Using the half-angle trigonometric identities $\cos\theta = \dfrac{1 - \tan^2\dfrac{\theta}{2}}{1 + \tan^2\dfrac{\theta}{2}}$ and $\sin\theta = \dfrac{2\tan\dfrac{\theta}{2}}{1 + \tan^2\dfrac{\theta}{2}}$, we get:
$A^{-1}B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
Answer: $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
Hence, the correct answer is option: 'A'.