Question

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

Answer 1

Answer: (D)

None of the above

The value of $\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$ is
$=\left[\begin{array}{cc}1+{\tan }^2\theta & \tan \theta -\tan \theta \\ \tan \theta -\tan \theta & {\tan }^2\theta +1\end{array}\right]$
$=\left[\begin{array}{cc}{\sec }^2\theta & 0\\ 0& {\sec }^2\theta \end{array}\right]$
Therefore, $\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]=\left[\begin{array}{cc}{\sec }^2\theta & 0\\ 0& {\sec }^2\theta \end{array}\right]$
$\Rightarrow a={\sec }^2\theta ,b=0$