**Given Information:**

The given information is that sec = 13/5.

**To Prove:**

We need to prove that 2sin - 3cos / 4sin - 9cos = 3.

**Solution:**

To solve this problem, we will use the fundamental trigonometric identities and properties.

**Step 1:** Rewrite sec in terms of sin and cos

The reciprocal identity of sec is cos.

Therefore, sec = 1 / cos.

Given that sec = 13/5, we can rewrite it as 1 / cos = 13/5.

**Step 2:** Solve for cos

To solve for cos, we can cross multiply:

5 = 13cos.

Then we divide both sides by 13 to isolate cos:

cos = 5/13.

**Step 3:** Use the Pythagorean Identity

The Pythagorean identity states that sin^2 + cos^2 = 1.

We can substitute the value of cos we found into this equation:

sin^2 + (5/13)^2 = 1.

Multiplying both sides by 13^2 to eliminate the fraction:

169sin^2 + 25 = 169.

Subtracting 25 from both sides:

169sin^2 = 144.

Dividing both sides by 169:

sin^2 = 144/169.

Taking the square root of both sides:

sin = ±√(144/169).

Since sin is positive in the first and second quadrants, we take the positive square root:

sin = √(144/169).

**Step 4:** Simplify the expression

Now, we can substitute the values of sin and cos we found into the expression:

2sin - 3cos / 4sin - 9cos.

Substituting sin = √(144/169) and cos = 5/13:

2(√(144/169)) - 3(5/13) / 4(√(144/169)) - 9(5/13).

Simplifying the expression further:

(2√144 - 15) / (4√144 - 45).

Since √144 = 12, we can substitute it into the expression:

(2 * 12 - 15) / (4 * 12 - 45).

Simplifying further:

(24 - 15) / (48 - 45).

(9) / (3).

3.

Therefore, we have proved that 2sin - 3cos / 4sin - 9cos = 3.