Question

If $\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ then-

• A. $a=1,b=1$
• B. $a=\cos 2\theta ,b=\sin 2\theta$
• C. $a=\sin 2\theta ,b=\cos 2\theta$
• D. None of these

Answer 1

The correct option is B $a=\cos 2\theta ,b=\sin 2\theta$

$\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$ .....(1)
Let $A=\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]$
$|A|=1+{\tan }^2\theta$
$adjA=C^T={\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$A^{-1}=\frac{adjA}{|A|}$
$A^{-1}=\frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$
Substitute this value in (1), we get
$\frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$=\frac{1}{1+{{\tan }^2\theta }^{}}\left[\begin{array}{cc}1-{\tan }^2\theta & -2\tan \theta \\ 2\tan \theta & 1-{\tan }^2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }& -\frac{2\tan \theta }{1+{\tan }^2\theta }\\ \frac{2\tan \theta }{1+{\tan }^2\theta }& \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
$\Rightarrow a=\cos 2\theta ,b=\sin 2\theta$