Question : In a triangle $XYZ$ right-angled at $Y$, if $XY=2\sqrt{6}$ and $XZ-YZ=2$, then $\sec X+\tan X$ is:
Option 1: $\frac{1}{\sqrt{6}}$
Option 2: $\sqrt{6}$
Option 3: $2\sqrt{6}$
Option 4: $\frac{\sqrt{6}}{2}$
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Correct Answer: $\sqrt{6}$
Solution : Given: $XY=2\sqrt{6}$ and $XZ-YZ=2$ So, $XY^2+YZ^2=XZ^2$ ------------------------------ (1) ⇒ $XZ^2-YZ^2=XY^2=(2\sqrt{6})^2$ ⇒ $(XZ+YZ)(XZ-YZ)=24$ ⇒ $XZ+YZ=12$ (As $XZ-YZ=2$) -------------------- (2) Now, after solving both equations, we get, $XZ=7$ and $YZ=5$ Now, $\sec X+\tan X$ = $\frac{7}{(2\sqrt{6})} + \frac{5}{(2\sqrt{6})} $ = $\frac{12}{(2\sqrt{6})}$ = $\sqrt{6}$ Hence, the correct answer is $\sqrt{6}$.
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