Question

If \[ \begin{bmatrix} 1 & -\tan\frac{\theta}{2}
[4pt] \tan\frac{\theta}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan\frac{\theta}{2}
[4pt] -\tan\frac{\theta}{2} & 1 \end{bmatrix}^{-1} = ? \]

• A. \[ \begin{bmatrix} \cos\theta & -\sin\theta
  \sin\theta & \cos\theta \end{bmatrix} \]
• B. \[ \begin{bmatrix} \cos\theta & \sin\theta
  -\sin\theta & \cos\theta \end{bmatrix} \]
• C. \[ \begin{bmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2}
  \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{bmatrix} \]
• D. \[ \begin{bmatrix} \sin\frac{\theta}{2} & \cos\frac{\theta}{2}
  -\cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \end{bmatrix} \] \bigskip

Hint:

Whenever matrices contain \(\tan\frac{\theta}{2}\), compute the matrix product first and then apply the identities \[ \cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}, \qquad \sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}. \] This often converts the result into a standard rotation matrix.

Answer 1

The Correct Option is A

Concept: For a matrix \[ A= \begin{bmatrix} a & b c & d \end{bmatrix}, \] its inverse is \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b -c & a \end{bmatrix}. \] Also, the half-angle identities are \[ \cos\theta= \frac{1-\tan^2\frac{\theta}{2}} {1+\tan^2\frac{\theta}{2}}, \] \[ \sin\theta= \frac{2\tan\frac{\theta}{2}} {1+\tan^2\frac{\theta}{2}}. \]

Step 1: Let \(t=\tan\frac{\theta}{2}\). Then the given expression becomes \[ \begin{bmatrix} 1 & -t t & 1 \end{bmatrix} \begin{bmatrix} 1 & t -t & 1 \end{bmatrix}^{-1}. \] Let \[ B= \begin{bmatrix} 1 & t -t & 1 \end{bmatrix}. \]

Step 2: Find \(B^{-1}\). The determinant of \(B\) is \[ |B|=1+t^2. \] Therefore, \[ B^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t t & 1 \end{bmatrix}. \]

Step 3: Multiply the matrices. Hence, \[ \begin{bmatrix} 1 & -t t & 1 \end{bmatrix} B^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t t & 1 \end{bmatrix}. \] Now, \[ = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t 2t & 1-t^2 \end{bmatrix}. \]

Step 4: Use half-angle identities. Using \[ \cos\theta= \frac{1-t^2}{1+t^2}, \] and \[ \sin\theta= \frac{2t}{1+t^2}, \] we obtain \[ \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t 2t & 1-t^2 \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \sin\theta & \cos\theta \end{bmatrix}. \]

Step 5: Write the final answer. \[ \boxed{ \begin{bmatrix} \cos\theta & -\sin\theta \sin\theta & \cos\theta \end{bmatrix} } \]