Question

If $x=\sec \theta \left[\begin{array}{cc}\sec \theta & \tan \theta \\ -\tan \theta & \sec \theta \end{array}\right]-\tan \theta \left[\begin{array}{cc}\tan \theta & \sec \theta \\ -\sec \theta & \tan \theta \end{array}\right]$, then $x$ is

$\left(a\right)$ Null matrix
$\left(b\right)$ Identity matrix
$\left(c\right)$ Triangular matrix
$\left(d\right)$ None of these

Answer 1

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

$\Rightarrow x=\left[\begin{array}{cc}{\sec }^2\theta & \sec \theta .\tan \theta \\ -\sec \theta .\tan \theta & {\sec }^2\theta \end{array}\right]-\left[\begin{array}{cc}{\tan }^2\theta & \sec \theta .\tan \theta \\ -\sec \theta .\tan \theta & {\tan }^2\theta \end{array}\right]$

$\Rightarrow x=\left[\begin{array}{cc}{\sec }^2\theta -{\tan }^2\theta & 0\\ 0& {\sec }^2\theta -{\tan }^2\theta \end{array}\right]$

$\Rightarrow x=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ which is an identity matrix.

$\therefore$ Option $\left(b\right)$ is correct.