1. Tardigrade
  2. Question
  3. Mathematics
  4. If beginpmatrix1 - tan θ tan θ 1 endpmatrix beginpmatrix1 tan θ - tan θ 1 endpmatrix-1= beginpmatrixa -b b a endpmatrix, then

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Q. If $\begin{pmatrix}1 & -\tan \theta \\ \tan \theta & 1\end{pmatrix}\begin{pmatrix}1 & \tan \theta \\ -\tan \theta & 1\end{pmatrix}^{-1}=\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$, then

Matrices

Solution:

$\begin{pmatrix}a & -b \\ b & a \end{pmatrix} =\begin{pmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{pmatrix}\begin{pmatrix} \cos ^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \cos ^2 \theta \end{pmatrix}$
$ =\begin{pmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{pmatrix}$
$ \Rightarrow a=\cos 2 \theta, b=\sin 2 \theta$