In a right-angled triangle XYZ right-angled at Y. if XY = 2 6 and XZ - YZ = 2, then secX + tanX is:

Answer: Option B

XZ - YZ = 2
h - P = 2 . . . . . . . (i)
h² = (2√6)² + P²
h² - P² = (2√6)²
(h - P)(h + P) = 4 × 6
(2)(h + P) = 24
h + P = 12 . . . . . . . (ii)
Adding equation (i) and (ii)
2h = 14
h = 7, P = 5
$$\begin{array}{rl}& \sec \text{X}+\tan \text{X}\\ & =\frac{\text{h}}{\text{XY}}+\frac{\text{P}}{\text{XY}}\\ & =\frac{7}{2\sqrt{6}}+\frac{5}{2\sqrt{6}}\\ & =\frac{12}{2\sqrt{6}}\\ & =\frac{6}{\sqrt{6}}\\ & =\sqrt{6}\end{array}$$