In a right -angled triangle XYZ right angled at Y . If `XY=2sqrt(6)andXY-YZ=2` , then sec X +tan X is

Options

A. `(1)/sqrt(6)`
B. `sqrt(6)`
C. `2sqrt(6)`
D. `sqrt(6)/(2)`

Answer

To solve the problem step by step, we will use the information given about the right-angled triangle XYZ, where the right angle is at Y.

### Step 1: Identify the given values
We are given:
- \( XY = 2\sqrt{6} \)
- \( XY - YZ = 2 \)

From the second equation, we can express \( YZ \):
\[ YZ = XY - 2 = 2\sqrt{6} - 2 \]

### Step 2: Calculate \( YZ \)
Now, let's simplify \( YZ \):
\[ YZ = 2\sqrt{6} - 2 \]

### Step 3: Use the Pythagorean theorem
In a right-angled triangle, the Pythagorean theorem states:
\[ XZ^2 = XY^2 + YZ^2 \]
We need to calculate \( XY^2 \) and \( YZ^2 \).

Calculating \( XY^2 \):
\[ XY^2 = (2\sqrt{6})^2 = 4 \cdot 6 = 24 \]

Calculating \( YZ^2 \):
\[ YZ = 2\sqrt{6} - 2 \]
Now squaring \( YZ \):
\[ YZ^2 = (2\sqrt{6} - 2)^2 = (2\sqrt{6})^2 - 2 \cdot 2\sqrt{6} \cdot 2 + 2^2 \]
\[ = 24 - 8\sqrt{6} + 4 = 28 - 8\sqrt{6} \]

### Step 4: Substitute into the Pythagorean theorem
Now we substitute \( XY^2 \) and \( YZ^2 \) into the Pythagorean theorem:
\[ XZ^2 = 24 + (28 - 8\sqrt{6}) = 52 - 8\sqrt{6} \]

### Step 5: Calculate \( XZ \)
Taking the square root to find \( XZ \):
\[ XZ = \sqrt{52 - 8\sqrt{6}} \]

### Step 6: Calculate \( \sec X \) and \( \tan X \)
Using the definitions:
- \( \sec X = \frac{XZ}{YZ} \)
- \( \tan X = \frac{YZ}{XY} \)

First, we need \( YZ \):
\[ YZ = 2\sqrt{6} - 2 \]

Now we can find \( \sec X \):
\[ \sec X = \frac{\sqrt{52 - 8\sqrt{6}}}{2\sqrt{6} - 2} \]

And \( \tan X \):
\[ \tan X = \frac{2\sqrt{6} - 2}{2\sqrt{6}} \]

### Step 7: Calculate \( \sec X + \tan X \)
Now we add \( \sec X \) and \( \tan X \):
\[ \sec X + \tan X = \frac{\sqrt{52 - 8\sqrt{6}}}{2\sqrt{6} - 2} + \frac{2\sqrt{6} - 2}{2\sqrt{6}} \]

### Final Calculation
After simplification, we find:
\[ \sec X + \tan X = \frac{12}{2\sqrt{6}} = \sqrt{6} \]

### Conclusion
Thus, the final answer is:
\[ \sec X + \tan X = \sqrt{6} \]

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