\( \cos \theta \frac{5}{13} \) and \( \sin \theta<0 \) Use a sketch to determine a) \( \sin \theta \) b) \( \sec \theta+\tan ^{2} \theta+1 \)

We are given that \[ \cos \theta = \frac{5}{13} \quad \text{and} \quad \sin \theta < 0. \] Since cosine is positive and sine is negative, the angle \(\theta\) is in the fourth quadrant. **Step 1. Compute \(\sin \theta\):** Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1, \] we substitute \(\cos \theta=\frac{5}{13}\): \[ \sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1 \quad \Rightarrow \quad \sin^2 \theta + \frac{25}{169} = 1. \] Solving for \(\sin^2 \theta\): \[ \sin^2 \theta = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}. \] Taking the square root and noting \(\sin \theta < 0\): \[ \sin \theta = -\frac{12}{13}. \] **Step 2. Compute \(\sec \theta + \tan^2 \theta + 1\):** First, find \(\sec \theta\) and \(\tan \theta\). - The secant is the reciprocal of the cosine: \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}. \] - The tangent is given by: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5}. \] Thus, \[ \tan^2 \theta = \left(-\frac{12}{5}\right)^2 = \frac{144}{25}. \] Now, sum the three terms: \[ \sec \theta + \tan^2 \theta + 1 = \frac{13}{5} + \frac{144}{25} + 1. \] Convert all terms to have a common denominator of 25: \[ \frac{13}{5} = \frac{65}{25}, \quad 1 = \frac{25}{25}. \] Thus, \[ \sec \theta + \tan^2 \theta + 1 = \frac{65}{25} + \frac{144}{25} + \frac{25}{25} = \frac{65 + 144 + 25}{25} = \frac{234}{25}. \] **Final Answers:** a) \(\sin \theta = -\frac{12}{13}\) b) \(\sec \theta + \tan^2 \theta + 1 = \frac{234}{25}\)