In a right -angled triangle ABC, right-angled at B, if `AB=2sqrt6 and AC-BC=2`, then find ` sec A+ tan A`.

Answer

To solve the problem step by step, we will follow the given information and apply the necessary mathematical concepts.

### Step 1: Understand the triangle and given information
We have a right-angled triangle ABC with the right angle at B. The lengths of the sides are given as follows:
- \( AB = 2\sqrt{6} \)
- \( AC - BC = 2 \)

### Step 2: Assign variables
Let \( BC = x \). According to the information provided, we can express \( AC \) in terms of \( x \):
\[ AC = x + 2 \]

### Step 3: Apply the Pythagorean theorem
In a right-angled triangle, the Pythagorean theorem states:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the known values:
\[ (2\sqrt{6})^2 + x^2 = (x + 2)^2 \]

### Step 4: Calculate \( AB^2 \)
Calculating \( (2\sqrt{6})^2 \):
\[ (2\sqrt{6})^2 = 4 \cdot 6 = 24 \]
So, we have:
\[ 24 + x^2 = (x + 2)^2 \]

### Step 5: Expand the right side
Now, expand \( (x + 2)^2 \):
\[ (x + 2)^2 = x^2 + 4x + 4 \]
Thus, the equation becomes:
\[ 24 + x^2 = x^2 + 4x + 4 \]

### Step 6: Simplify the equation
Subtract \( x^2 \) from both sides:
\[ 24 = 4x + 4 \]
Now, subtract 4 from both sides:
\[ 20 = 4x \]
Dividing both sides by 4 gives:
\[ x = 5 \]

### Step 7: Find \( AC \)
Now that we have \( BC = x = 5 \), we can find \( AC \):
\[ AC = x + 2 = 5 + 2 = 7 \]

### Step 8: Find \( \sec A + \tan A \)
We know:
- \( AB = 2\sqrt{6} \) (adjacent side)
- \( BC = 5 \) (opposite side)
- \( AC = 7 \) (hypotenuse)

Now, we can find \( \sec A \) and \( \tan A \):
\[ \sec A = \frac{AC}{AB} = \frac{7}{2\sqrt{6}} \]
\[ \tan A = \frac{BC}{AB} = \frac{5}{2\sqrt{6}} \]

### Step 9: Calculate \( \sec A + \tan A \)
Adding the two values together:
\[ \sec A + \tan A = \frac{7}{2\sqrt{6}} + \frac{5}{2\sqrt{6}} = \frac{7 + 5}{2\sqrt{6}} = \frac{12}{2\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \]

### Final Answer
Thus, the value of \( \sec A + \tan A \) is:
\[ \sqrt{6} \]