If \( \sec \theta=\frac{13}{12} \), find the values of other trigonometric ratios.

Given that \( \sec \theta = \frac{13}{12} \), we can find the other trigonometric ratios using the relationship between secant and cosine. Recall that: \[ \sec \theta = \frac{1}{\cos \theta} \] From this, we can find \( \cos \theta \): \[ \cos \theta = \frac{1}{\sec \theta} = \frac{12}{13} \] Next, we can find \( \sin \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \cos \theta \): \[ \sin^2 \theta + \left(\frac{12}{13}\right)^2 = 1 \] Calculating \( \left(\frac{12}{13}\right)^2 \): \[ \left(\frac{12}{13}\right)^2 = \frac{144}{169} \] Now substituting this back into the identity: \[ \sin^2 \theta + \frac{144}{169} = 1 \] To isolate \( \sin^2 \theta \), we subtract \( \frac{144}{169} \) from both sides: \[ \sin^2 \theta = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] Taking the square root to find \( \sin \theta \): \[ \sin \theta = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \] The sign of \( \sin \theta \) depends on the quadrant in which \( \theta \) lies. For now, we will consider both possibilities. Next, we can find the other trigonometric ratios: 1. **Tangent**: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\pm \frac{5}{13}}{\frac{12}{13}} = \pm \frac{5}{12} \] 2. **Cosecant**: \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\pm \frac{5}{13}} = \pm \frac{13}{5} \] 3. **Cotangent**: \[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\pm \frac{5}{12}} = \pm \frac{12}{5} \] Now, summarizing all the trigonometric ratios: - \( \sec \theta = \frac{13}{12} \) - \( \cos \theta = \frac{12}{13} \) - \( \sin \theta = \pm \frac{5}{13} \) - \( \tan \theta = \pm \frac{5}{12} \) - \( \csc \theta = \pm \frac{13}{5} \) - \( \cot \theta = \pm \frac{12}{5} \) Thus, the values of the other trigonometric ratios are: \[ \sin \theta = \pm \frac{5}{13}, \quad \tan \theta = \pm \frac{5}{12}, \quad \csc \theta = \pm \frac{13}{5}, \quad \cot \theta = \pm \frac{12}{5} \]