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  4. If [1&- tanθ tan θ&1][1& tan θ - tan θ &1]-1 = [a&-b b&a] then

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

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Solution:

$\begin{bmatrix}1&- \tan\theta \\ \tan \theta&1\end{bmatrix}$
$\begin{bmatrix}1&\tan \theta \\ - \tan \theta &1\end{bmatrix}^{-1} = \begin{bmatrix}a&-b\\ b&a\end{bmatrix}$
Since $ A^{-1} = \frac{adj A}{\left|A\right|}$
$\therefore \ \begin{bmatrix}1&- \tan \theta \\ \tan \theta &1\end{bmatrix} $
$\frac{\begin{bmatrix}1&- \tan \theta \\ \tan \theta &1\end{bmatrix}}{1+\tan^{2} \theta} = \begin{bmatrix}a&-b\\ b&a\end{bmatrix} $
$\frac{1}{1+\tan ^{2} \theta } \begin{bmatrix}1-\tan^{2} \theta &- \tan \theta -\tan \theta \\ \tan \theta +\tan \theta &-\tan^{2} \theta+ 1\end{bmatrix} $
$ = \begin{bmatrix}a&-b\\ b&a\end{bmatrix} $
$\begin{bmatrix}\frac{1-\tan ^{2} \theta }{1+ \tan^{2} \theta } &\frac{-2 \tan \theta }{1+ \tan^{2} \theta } \\ \frac{2 \tan\theta}{1+ \tan^{2} \theta } &\frac{1 -\tan ^{2} \theta}{1+\tan^{2} \theta } \end{bmatrix} = \begin{bmatrix}a&-b\\ b&a\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}\cos2\theta&-\sin2\theta\\ \sin 2\theta &\cos 2\theta \end{bmatrix} = \begin{bmatrix}a&-b\\ b&a\end{bmatrix} $
So, $a = \cos 2\theta , b = \sin 2\theta $