Question

If $I=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ , then $\left(a^2+b^2\right)$ is equal to $\dots \dots$

Answer 1

Answer: 1

Here, it is given that
$I=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}$
$=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\times \frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$=\frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1-{\tan }^2\theta & -2\tan \theta \\ 2\tan \theta & 1-{\tan }^2\theta \end{array}\right]$
$=\left[\begin{array}{cc}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }& \frac{-2\tan \theta }{1+{\tan }^2\theta }\\ \frac{2\tan \theta }{1+{\tan }^2\theta }& \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\end{array}\right]$
$=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
since, $I=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$\therefore {\cos }^{2}2\theta +{\sin }^{2}2\theta =1$