Find the exact value of each of the following under the given conditions below. \[ \tan \alpha=-\frac{4}{3}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{\sqrt{3}}{2}, 0<\beta<\frac{\pi}{2} \] \( \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array} \)

To find the exact values of \( \sin(\alpha + \beta) \), \( \cos(\alpha + \beta) \), \( \sin(\alpha - \beta) \), and \( \tan(\alpha - \beta) \) under the provided conditions, we'll use the sine and cosine addition formulas, along with the Pythagorean identity. Given: - \( \tan \alpha = -\frac{4}{3} \) which means in the second quadrant, where sine is positive and cosine is negative. - \( \cos \beta = \frac{\sqrt{3}}{2} \) which means \( \beta = \frac{\pi}{6} \) since \( 0 < \beta < \frac{\pi}{2} \). ### Step 1: Finding \( \sin \alpha \) and \( \cos \alpha \) From \( \tan \alpha = -\frac{4}{3} \), we can use: \[ \sin \alpha = -\frac{4}{\sqrt{(4^2)+(3^2)}} = -\frac{4}{5}, \quad \cos \alpha = -\frac{3}{\sqrt{(4^2)+(3^2)}} = -\frac{3}{5} \] Since \( \alpha \) is in the second quadrant, we take: \[ \sin \alpha = \frac{4}{5}, \quad \cos \alpha = -\frac{3}{5} \] ### Step 2: Finding \( \sin \beta \) Since \( \beta = \frac{\pi}{6} \): \[ \sin \beta = \frac{1}{2} \] ### Step 3: Calculating the expressions #### (a) \( \sin(\alpha + \beta) \) Using the sine addition formula: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] \[ = \left(\frac{4}{5}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{3}{5}\right)\left(\frac{1}{2}\right) \] \[ = \frac{4\sqrt{3}}{10} - \frac{3}{10} = \frac{4\sqrt{3} - 3}{10} \] #### (b) \( \cos(\alpha + \beta) \) Using the cosine addition formula: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] \[ = \left(-\frac{3}{5}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{4}{5}\right)\left(\frac{1}{2}\right) \] \[ = -\frac{3\sqrt{3}}{10} - \frac{4}{10} = -\frac{3\sqrt{3} + 4}{10} \] #### (c) \( \sin(\alpha - \beta) \) Using the sine subtraction formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] \[ = \left(\frac{4}{5}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{3}{5}\right)\left(\frac{1}{2}\right) \] \[ = \frac{4\sqrt{3}}{10} + \frac{3}{10} = \frac{4\sqrt{3} + 3}{10} \] #### (d) \( \tan(\alpha - \beta) \) Using the tangent subtraction formula: \[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} \] We need \( \cos(\alpha - \beta) \): \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] \[ = \left(-\frac{3}{5}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{2}\right) \] \[ = -\frac{3\sqrt{3}}{10} + \frac{4}{10} = \frac{4 - 3\sqrt{3}}{10} \] Now compute