Write the area of the triangle formed by the coordinate axes and the line (sec θ − tan θ) *x* + (sec θ + tan θ) y = 2.

#### Solution

The point of intersection of the coordinate axes is (0, 0).

Let us find the intersection of the line (sec θ − tan θ) x + (sec θ + tan θ) y = 2 and the coordinate axis.

For x-axis:

y = 0, \[x = \frac{2}{sec\theta - tan\theta}\]

For y-axis:

x = 0,

\[y = \frac{2}{sec\theta + tan\theta}\]

Thus, the coordinates of the triangle formed by the coordinate axis and the line (sec θ − tan θ) *x* + (sec θ + tan θ) *y* = 2 are (0, 0), \[\left( \frac{2}{sec\theta - tan\theta}, 0 \right)\] and \[\left( 0, \frac{2}{sec\theta + tan\theta} \right)\].

Let A be the area of the required triangle..

\[\therefore A = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ \frac{2}{\sec\theta - tan\theta} & 0 & 1 \\ 0 & \frac{2}{\sec\theta + tan\theta} & 1\end{vmatrix}\]

\[ \Rightarrow A = \frac{1}{2} \times \frac{2}{\sec\theta - tan\theta} \times \frac{2}{\sec\theta + tan\theta}\]

\[ \Rightarrow A = \frac{2}{\left( \sec\theta - tan\theta \right)\left( \sec\theta + tan\theta \right)} = \frac{2}{\left( \sec^2 \theta - \tan^2 \theta \right)} = 2\]

Hence, the area of the triangle is 2 square units.