Question

If $A\left(\theta \right)=\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]$ and $AB=I$, then $\left({\sec }^2\theta \right)B$ is equal to

• A. $A\left(\theta \right)$
• B. $A(-\theta )$
• C. $A\left(\frac{\theta }{2}\right)$
• D. $A\left(\frac{-\theta }{2}\right)$

Answer 1

The correct option is B $A(-\theta )$

$A\left(\theta \right)=\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]$ and $AB=I$
As $AB=I$, we get $B=A^{-1}$----i

$adjA=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$

$A^{-1}=\frac{adjA}{|A|}$

$\therefore B=\frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$------from i
$\Rightarrow \left({\sec }^2\theta \right)B=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]=A(-\theta )$
Hence, option B.