If \( \cos \theta=\frac{9}{15} \), then find \( \tan \theta, \sec \theta \) and \( \sin \theta \) If \( \operatorname{cosec} \theta=\frac{17}{8} \), then find \( \sin \theta \) and \( \cos \theta \)

First, let's simplify \( \cos \theta = \frac{9}{15} \). This reduces to \( \cos \theta = \frac{3}{5} \). Using the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), we find: \[ \sin^2 \theta = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] Thus, \( \sin \theta = \frac{4}{5} \) (we take the positive root for angles in the first quadrant). Next, we can calculate \( \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \] Now, let's find \( \sec \theta \): \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \] Now, for the second part where \( \operatorname{cosec} \theta = \frac{17}{8} \), we know that: \[ \sin \theta = \frac{1}{\operatorname{cosec} \theta} = \frac{1}{\frac{17}{8}} = \frac{8}{17} \] Using the Pythagorean identity again to find \( \cos \theta \): \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left( \frac{8}{17} \right)^2 = 1 - \frac{64}{289} = \frac{225}{289} \] Thus, \( \cos \theta = \frac{15}{17} \) (again, taking the positive root for angles in the first quadrant). In conclusion, we have: - \( \tan \theta = \frac{4}{3} \) - \( \sec \theta = \frac{5}{3} \) - \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = \frac{3}{5} \) for the first part, - \( \sin \theta = \frac{8}{17} \) and \( \cos \theta = \frac{15}{17} \) for the second part.