If a right-angled triangle XYZ right-angled at Y. If XY = $${\text{2}}\sqrt 6 $$  and XZ - YZ = 2, then secX + tanX is?

Solution(By Examveda Team)

$$\eqalign{ & XZ - YZ = 2 \cr & h - P = 2 \,........{\text{(i)}} \cr & {h^2} = {(2\sqrt 6 )^2} + {P^2} \cr & {h^2} - {P^2} = {\left( {2\sqrt 6 } \right)^2} \cr & (h - P)(h + P) = 4 \times 6 \cr & (2)(h + P) = 24 \cr & h + P = 12 \,..........{\text{(ii)}} \cr & {\text{Adding eq }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 2h = 14 \cr & h = 7 \,{\text{and }}P = 5 \cr & \therefore secX + tanX \cr & = \frac{h}{{XY}} + \frac{P}{{XY}} \cr & = \frac{7}{{2\sqrt 6 }} + \frac{5}{{2\sqrt 6 }} \cr & = \frac{{12}}{{2\sqrt 6 }} \cr & = \frac{6}{{\sqrt 6 }} \cr & = \sqrt 6 \cr} $$