Question

The determinant $\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 12& 10& 2\end{array}\right|$ equals

• A. $2{\sin }^2\theta$
• B. $12{\sec }^2\theta -10{\tan }^2\theta$
• C. $12{\sec }^2\theta -10{\tan }^2\theta +5$
• D. 0

Answer 1

The correct option is D 0

$\Delta =\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 12& 10& 2\end{array}\right|$
Applying $C_2\to C_2-C_1$
$\Delta =\left|\begin{array}{ccc}{\sec }^2\theta & {\sec }^2\theta -{\tan }^2\theta & 1\\ {\tan }^2\theta & {\tan }^2\theta -{\sec }^2\theta & -1\\ 12& 12-10& 2\end{array}\right|=\left|\begin{array}{ccc}{\sec }^2\theta & 1& 1\\ {\tan }^2\theta & -1& -1\\ 12& 2& 2\end{array}\right|=0$
Hence, option 'D' is correct.