Mathematics - Matrix Question with Solution | TestHub

Given, $\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ $\Rightarrow \left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\cdot \frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$ $=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ $\Rightarrow \frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1-{\tan }^2\theta & -2\tan \theta \\ 2\tan \theta & 1-{\tan }^2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ $\Rightarrow \left[\begin{array}{cc}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }& \frac{2\tan \theta }{1+{\tan }^2\theta }\\ \frac{2\tan \theta }{1+{\tan }^2\theta }& \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ $\Rightarrow \left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ $\Rightarrow a=\cos 2\theta ,b=\sin 2\theta$