1. **State the problem:** We want to use trigonometric substitution to express the algebraic expression $$\sqrt{25x^2 + 9}$$ as a function of $$\sec(\theta)$$, given the substitution $$5x = 3\tan(\theta)$$ and the domain $$0 \leq \theta \leq \frac{\pi}{2}$$. 2. **Recall the substitution and identity:** Given $$5x = 3\tan(\theta)$$, we can write $$x = \frac{3}{5}\tan(\theta)$$. 3. **Rewrite the expression inside the square root:** $$25x^2 + 9 = 25\left(\frac{3}{5}\tan(\theta)\right)^2 + 9 = 25 \cdot \frac{9}{25} \tan^2(\theta) + 9 = 9\tan^2(\theta) + 9$$ 4. **Factor out 9:** $$9\tan^2(\theta) + 9 = 9(\tan^2(\theta) + 1)$$ 5. **Use the Pythagorean identity:** Recall that $$1 + \tan^2(\theta) = \sec^2(\theta)$$, so $$9(\tan^2(\theta) + 1) = 9\sec^2(\theta)$$ 6. **Take the square root:** $$\sqrt{25x^2 + 9} = \sqrt{9\sec^2(\theta)} = 3|\sec(\theta)|$$ 7. **Determine the sign of $$\sec(\theta)$$:** Since $$0 \leq \theta \leq \frac{\pi}{2}$$, $$\sec(\theta) = \frac{1}{\cos(\theta)}$$ is positive, so $$|\sec(\theta)| = \sec(\theta)$$ 8. **Final expression:** $$\sqrt{25x^2 + 9} = 3\sec(\theta)$$ This expresses the original algebraic expression as a function of $$\sec(\theta)$$ using the given substitution.