# The value of the determinant

Question:

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

Solution:

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

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

Applying $C_{2} \rightarrow C_{2}+C_{3}$

$\Rightarrow \Delta=\left|\begin{array}{ccc}\sec ^{2} \theta & \tan ^{2} \theta+1 & 1 \\ \tan ^{2} \theta & \sec ^{2} \theta-1 & -1 \\ 22 & 20+2 & 2\end{array}\right|$

$\Rightarrow \Delta=\left|\begin{array}{ccc}\sec ^{2} \theta & \sec ^{2} \theta & 1 \\ \tan ^{2} \theta & \tan ^{2} \theta & -1 \\ 22 & 22 & 2\end{array}\right|$

$\Rightarrow \Delta=0$        ( $\because$ The value of determinant with two identicals columns is equal to zero)

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