In triangle XYZ, ∠Y is a right angle. Given: XY = \(2\sqrt{6}\) and XZ − YZ = 2 Find the value of sec X + tan X.

Q.In triangle XYZ, ∠Y is a right angle. Given: XY = \(2\sqrt{6}\) and XZ − YZ = 2 Find the value of sec X + tan X.

In triangle ∆XYZ, ∠Y is a right angle. From the diagram, we get: \[ \sec X - \tan X = \frac{XZ}{XY} - \frac{YZ}{XY} = \frac{XZ - YZ}{XY} = \frac{2}{2\sqrt{6}} \quad \text{[Substituting XZ − YZ = 2 and XY = 2√6]} = \frac{1}{\sqrt{6}} \] We know: \[ \sec^2 X - \tan^2 X = 1 \Rightarrow (\sec X + \tan X)(\sec X - \tan X) = 1 \Rightarrow (\sec X + \tan X) \cdot \frac{1}{\sqrt{6}} = 1 \Rightarrow \sec X + \tan X = \sqrt{6} \] Answer: \[ \sec X + \tan X = \sqrt{6} \](Answer)