Question

Prove that ${\displaystyle \left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 38& 36& 2\end{array}\right|}$ = 0

Answer 1

Let Δ = ${\displaystyle \left|\begin{array}{ccc}{\sec }^2\theta & {\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta & -1\\ 38& 36& 2\end{array}\right|}$

Δ = ${\displaystyle \left|\begin{array}{ccc}{\sec }^2\theta & 1+{\tan }^2\theta & 1\\ {\tan }^2\theta & {\sec }^2\theta -1& -1\\ 38& 38& 2\end{array}\right| {\text{C}}_2\to {\text{C}}_2+{\text{C}}_3}$

Δ = ${\displaystyle \left|\begin{array}{ccc}{\sec }^2\theta & {\sec }^2\theta & 1\\ {\tan }^2\theta & {\tan }^2\theta & -1\\ 38& 38& 2\end{array}\right|}$

${\displaystyle {\sec }^2\theta -{\tan }^2\theta }$ = 1

${\displaystyle {\sec }^2\theta =1+{\tan }^2\theta }$

${\displaystyle {\sec }^2\theta -1={\tan }^2\theta }$

Δ = 0

Two columns are same.