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Question: \(\left[ \begin{array} { c c } 1 & - \tan \theta / 2 \\ \tan \theta / 2 & 1 \end{array} \right]\) \(...

[1−tan⁡θ/2tan⁡θ/21]\left[ \begin{array} { c c } 1 & - \tan \theta / 2 \\ \tan \theta / 2 & 1 \end{array} \right] $\begin{bmatrix} 1 & \tan\theta/2 \

  • \tan\theta/2 & 1 \end{bmatrix}^{- 1}$ is equal to-
A

[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\begin{bmatrix} \sin\theta & - \cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}

B

[cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

C

$\begin{bmatrix} \cos\theta & \sin\theta \

  • \sin\theta & \cos\theta \end{bmatrix}$
D

None of these

Answer

[cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

Explanation

Solution

$\begin{bmatrix} 1 & \tan\theta/2 \

  • \tan\theta/2 & 1 \end{bmatrix}^{- 1}== \frac{\begin{bmatrix} 1 & - \tan\theta/2 \ \tan\theta/2 & 1 \end{bmatrix}}{1 + \tan^{2}\theta/2}$

= [1−tan⁡θ/2tan⁡θ/21]\begin{bmatrix} 1 & - \tan\theta/2 \\ \tan\theta/2 & 1 \end{bmatrix} [1−tan⁡θ/2tan⁡θ/21]1+tan⁡2θ/2\frac{\begin{bmatrix} 1 & - \tan\theta/2 \\ \tan\theta/2 & 1 \end{bmatrix}}{1 + \tan^{2}\theta/2}

= [1−tan⁡2θ/21+tan⁡2θ/2−2tan⁡θ/21+tan⁡2θ/22tan⁡θ/21+tan⁡2θ/21−tan⁡2θ/21+tan⁡2θ/2]\begin{bmatrix} \frac{1 - \tan^{2}\theta/2}{1 + \tan^{2}\theta/2} & \frac{- 2\tan\theta/2}{1 + \tan^{2}\theta/2} \\ \frac{2\tan\theta/2}{1 + \tan^{2}\theta/2} & \frac{1 - \tan^{2}\theta/2}{1 + \tan^{2}\theta/2} \end{bmatrix}= [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}