Question

The value of the determinant $∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 22& 20& 2\end{array}\right|$ is _____________.

Answer 1

Given: $∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 22& 20& 2\end{array}\right|$

$$\begin{gathered} ∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 22& 20& 2\end{array}\right| \\ \mathrm{Applying}{C}_2\to C_2+C_3 \\ \Rightarrow ∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta +1& 1\\ {\tan }^2\theta & {\sec }^2\theta -1& -1\\ 22& 20+2& 2\end{array}\right| \\ \Rightarrow ∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\sec }^2\theta & 1\\ {\tan }^2\theta & {\tan }^2\theta & -1\\ 22& 22& 2\end{array}\right| \\ \Rightarrow ∆=0\left(\because \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{with}\mathrm{two}\mathrm{identicals}\mathrm{columns}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{zero}\right) \end{gathered}$$

Hence, the value of the determinant $∆=\left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 22& 20& 2\end{array}\right|$ is 0.