If \( \tan (\theta)=\frac{4}{3}, 0 \leq \theta \leq \frac{\pi}{2} \), then \( \sin (\theta) \) equals \( \cos (\theta) \) equals \( \sec (\theta) \) equals \( \square \)

Given that \( \tan(\theta) = \frac{4}{3} \), we can use the definition of tangent to find the sine and cosine values. Recall that: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] From the given information, we can set: \[ \sin(\theta) = 4k \quad \text{and} \quad \cos(\theta) = 3k \] for some positive constant \( k \). Now, we can use the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Substituting our expressions for sine and cosine: \[ (4k)^2 + (3k)^2 = 1 \] This simplifies to: \[ 16k^2 + 9k^2 = 1 \] \[ 25k^2 = 1 \] \[ k^2 = \frac{1}{25} \] \[ k = \frac{1}{5} \] Now we can find \( \sin(\theta) \) and \( \cos(\theta) \): \[ \sin(\theta) = 4k = 4 \cdot \frac{1}{5} = \frac{4}{5} \] \[ \cos(\theta) = 3k = 3 \cdot \frac{1}{5} = \frac{3}{5} \] Next, we can find \( \sec(\theta) \), which is the reciprocal of \( \cos(\theta) \): \[ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \] Thus, we have: \[ \sin(\theta) = \frac{4}{5}, \quad \cos(\theta) = \frac{3}{5}, \quad \sec(\theta) = \frac{5}{3} \] In summary: \[ \sin(\theta) = \frac{4}{5}, \quad \cos(\theta) = \frac{3}{5}, \quad \sec(\theta) = \frac{5}{3} \]