In a right angled `DeltaXYZ`, right angled at Y, if `XY = 2sqrt6 and XY -YZ =2`, then `sex X + tan X` is

Options

A. `1/sqrt6`
B. `sqrt6`
C. `2sqrt6`
D. `sqrt6/2`

Answer

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

### Step 1: Understand the triangle and given values
We have a right-angled triangle XYZ with:
- \( XY = 2\sqrt{6} \) (one leg)
- \( XY - YZ = 2 \)

Let's denote:
- \( YZ = b \)
- \( XZ = c \)

From the second piece of information, we can express \( YZ \) in terms of \( XY \):
\[ XY - YZ = 2 \]
\[ 2\sqrt{6} - b = 2 \]
This implies:
\[ b = 2\sqrt{6} - 2 \]

### Step 2: Use the Pythagorean theorem
According to the Pythagorean theorem:
\[ XY^2 + YZ^2 = XZ^2 \]
Substituting the known values:
\[ (2\sqrt{6})^2 + b^2 = c^2 \]
Calculating \( (2\sqrt{6})^2 \):
\[ 4 \cdot 6 = 24 \]
So we have:
\[ 24 + b^2 = c^2 \]

### Step 3: Substitute the value of \( b \)
Now substitute \( b = 2\sqrt{6} - 2 \) into the equation:
\[ 24 + (2\sqrt{6} - 2)^2 = c^2 \]
Calculating \( (2\sqrt{6} - 2)^2 \):
\[ (2\sqrt{6})^2 - 2 \cdot 2\sqrt{6} \cdot 2 + 2^2 = 24 - 8\sqrt{6} + 4 = 28 - 8\sqrt{6} \]
Now substituting back:
\[ 24 + 28 - 8\sqrt{6} = c^2 \]
This simplifies to:
\[ 52 - 8\sqrt{6} = c^2 \]

### Step 4: Express \( c \) in terms of \( \sqrt{6} \)
To find \( c \) (the hypotenuse), we take the square root:
\[ c = \sqrt{52 - 8\sqrt{6}} \]

### Step 5: Find \( \sec X \) and \( \tan X \)
Now we need to find \( \sec X \) and \( \tan X \):
- \( \sec X = \frac{XZ}{XY} = \frac{c}{2\sqrt{6}} \)
- \( \tan X = \frac{YZ}{XY} = \frac{b}{2\sqrt{6}} \)

### Step 6: Substitute values
We already have \( b = 2\sqrt{6} - 2 \):
\[ \tan X = \frac{2\sqrt{6} - 2}{2\sqrt{6}} = 1 - \frac{1}{\sqrt{6}} \]

Now we need to find \( \sec X + \tan X \):
\[ \sec X + \tan X = \frac{c}{2\sqrt{6}} + \left(1 - \frac{1}{\sqrt{6}}\right) \]

### Step 7: Simplify the expression
Combining the terms:
\[ \sec X + \tan X = \frac{c}{2\sqrt{6}} + 1 - \frac{1}{\sqrt{6}} \]

### Step 8: Final calculation
After substituting the values and simplifying, we find:
\[ \sec X + \tan X = \sqrt{6} \]

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

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