2. If \( \sin \alpha-\frac{3}{5} \) with \( \alpha \in\left[90^{\circ}, 270^{\circ}\right] \) and \( \cos \beta=-\frac{12}{13} \). wht \( \beta \in\left[0^{\circ}, 10^{\circ}\right] \) Calcutcte withot the use क a calculatr and with the aid of a diagram. te value \& \( \cos \alpha+\tan \beta \).

To begin, let's analyze the equation \( \sin \alpha - \frac{3}{5} = 0 \), which gives us \( \sin \alpha = \frac{3}{5} \). Given that \( \alpha \) lies within the interval \( [90^{\circ}, 270^{\circ}]\), we can conclude that \( \alpha \) is in the second or third quadrant. This implies that \( \alpha = 180^{\circ} + \theta \) for some angle \( \theta \) such that \( \sin \theta = \frac{3}{5} \). To find \( \cos \alpha \), we use the Pythagorean identity: \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] Thus, \( \cos \alpha = -\frac{4}{5} \) because cosine is negative in the second and third quadrants. Next, let's analyze \( \beta \) where \( \cos \beta = -\frac{12}{13} \). Since \( \beta \) is in the interval \( [0^{\circ}, 90^{\circ}] \), this means \( \beta \) must be in the second quadrant, corresponding to \( \cos \beta\)'s negative value. Using the same Pythagorean identity, we find \( \sin \beta \): \[ \sin^2 \beta + \left(-\frac{12}{13}\right)^2 = 1 \implies \sin^2 \beta + \frac{144}{169} = 1 \implies \sin^2 \beta = 1 - \frac{144}{169} = \frac{25}{169} \] Thus, \( \sin \beta = \frac{5}{13} \). Now we can find \( \cos \alpha + \tan \beta \): We compute \( \tan \beta \) as follows: \[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12} \] Then we can find: \[ \cos \alpha + \tan \beta = -\frac{4}{5} - \frac{5}{12} \] To combine these fractions, we’ll need a common denominator: \[ -\frac{4}{5} = -\frac{48}{60}, \quad -\frac{5}{12} = -\frac{25}{60} \implies -\frac{48}{60} - \frac{25}{60} = -\frac{73}{60} \] Thus, the final answer is: \[ \cos \alpha + \tan \beta = -\frac{73}{60} \]