If [1−tan⁡θtan⁡θ1][1tan⁡θ−tan⁡θ1]−1=[a−bba]\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}…

Problem

If [1−tan⁡θtan⁡θ1][1tan⁡θ−tan⁡θ1]−1=[a−bba]\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} , then : (a) a=b=1a = b = 1 (b) a=cos⁡2θ,b=sin⁡2θa = \cos 2\theta, b = \sin 2\theta (c) a=sin⁡2θ,b=cos⁡2θa = \sin 2\theta, b = \cos 2\theta (d) a=1,b=sin⁡2θa = 1, b = \sin 2\theta

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