1. **State the problem:** We want to express the algebraic expressions $4 = 25x^2 + 9$ and $5x = 3\tan\theta$ as functions of $\sec\theta$ where $0 \leq \theta \leq \frac{\pi}{2}$. 2. **Given substitution:** We have $5x = 3\tan\theta$, so $x = \frac{3}{5} \tan\theta$. 3. **Rewrite the expression $4 = 25x^2 + 9$ using $x$ in terms of $\tan\theta$:** Substitute $x$: $$ 4 = 25 \left(\frac{3}{5} \tan\theta\right)^2 + 9 $$ Simplify inside the square: $$ 4 = 25 \times \frac{9}{25} \tan^2\theta + 9 = 9 \tan^2\theta + 9 $$ 4. **Factor the right side:** $$ 4 = 9 (\tan^2\theta + 1) $$ 5. **Use the Pythagorean identity:** Recall that $1 + \tan^2\theta = \sec^2\theta$. So, $$ 4 = 9 \sec^2\theta $$ 6. **Solve for $\sec^2\theta$:** $$ \sec^2\theta = \frac{4}{9} $$ 7. **Express $\sec\theta$:** Since $0 \leq \theta \leq \frac{\pi}{2}$, $\sec\theta$ is positive, so $$ \sec\theta = \frac{2}{3} $$ **Final answers:** - $x = \frac{3}{5} \tan\theta$ - $4 = 9 \sec^2\theta$ - $\sec\theta = \frac{2}{3}$