Q 1. Find \( \sin \theta \) and \( \tan \theta \), if \( \cos \theta=-\frac{12}{13} \) and \( \theta \) lies in the third quadrant.

To find \( \sin \theta \) and \( \tan \theta \) given that \( \cos \theta = -\frac{12}{13} \) and \( \theta \) is in the third quadrant, we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] First, we can calculate \( \sin^2 \theta \): \[ \sin^2 \theta + \left(-\frac{12}{13}\right)^2 = 1 \] Calculating \( \left(-\frac{12}{13}\right)^2 \): \[ \left(-\frac{12}{13}\right)^2 = \frac{144}{169} \] Now substituting this into the identity: \[ \sin^2 \theta + \frac{144}{169} = 1 \] To isolate \( \sin^2 \theta \), we subtract \( \frac{144}{169} \) from both sides: \[ \sin^2 \theta = 1 - \frac{144}{169} \] Converting 1 to a fraction with a denominator of 169: \[ 1 = \frac{169}{169} \] Now we can perform the subtraction: \[ \sin^2 \theta = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] Taking the square root of both sides gives us: \[ \sin \theta = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \] Since \( \theta \) is in the third quadrant, where sine is negative, we have: \[ \sin \theta = -\frac{5}{13} \] Next, we can find \( \tan \theta \) using the definition of tangent: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Substituting the values we found: \[ \tan \theta = \frac{-\frac{5}{13}}{-\frac{12}{13}} = \frac{-5}{13} \cdot \frac{13}{-12} = \frac{5}{12} \] Thus, the final results are: \[ \sin \theta = -\frac{5}{13}, \quad \tan \theta = \frac{5}{12} \]