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: Evaluate the inverse first and then identify the resulting matrix using half-angle identities.

Step 1: Let \(t=\tan\frac{\theta}{2}\). The inverse of \(\begin{bmatrix}1&t\\-t&1\end{bmatrix}\) is \(\dfrac1{1+t^2}\begin{bmatrix}1&-t\\t&1\end{bmatrix}\).

Step 2: Multiplying gives \(\dfrac1{1+t^2}\begin{bmatrix}1-t^2&-2t\\2t&1-t^2\end{bmatrix}\).

Step 3: Using \(\cos\theta=\dfrac{1-t^2}{1+t^2}\) and \(\sin\theta=\dfrac{2t}{1+t^2}\), the matrix becomes \(\boxed{\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}}\).