Find the exact value of the expressions \( \cos (\alpha+\beta) \), \( \sin (\alpha+\beta) \) and \( \tan (\alpha+\beta) \) under the following conditions: \( \cos (\alpha)=\frac{24}{25}, \alpha \) lies in quadrant IV, and \( \sin (\beta)=\frac{-2}{5}, \beta \) lies in quadrant Ill. a. \( \cos (\alpha+\beta)=\frac{-24 \sqrt{21}-14}{125} \) (Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression.) b. \( \sin (\alpha+\beta)=\frac{7 \sqrt{21}-48}{125} \) (Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression.) c. tan \( (\alpha+\beta)=\square \) (Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression.)

To find the exact values of \( \cos(\alpha+\beta) \), \( \sin(\alpha+\beta) \), and \( \tan(\alpha+\beta) \), we will use the angle addition formulas: 1. \( \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \) 2. \( \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \) 3. \( \tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} \) ### Step 1: Find \( \sin(\alpha) \) and \( \cos(\beta) \) Given: - \( \cos(\alpha) = \frac{24}{25} \) and \( \alpha \) is in quadrant IV. - \( \sin(\beta) = \frac{-2}{5} \) and \( \beta \) is in quadrant III. #### Finding \( \sin(\alpha) \): Using the Pythagorean identity: \[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \] \[ \sin^2(\alpha) + \left(\frac{24}{25}\right)^2 = 1 \] \[ \sin^2(\alpha) + \frac{576}{625} = 1 \] \[ \sin^2(\alpha) = 1 - \frac{576}{625} = \frac{49}{625} \] \[ \sin(\alpha) = -\sqrt{\frac{49}{625}} = -\frac{7}{25} \quad (\text{since } \alpha \text{ is in quadrant IV}) \] #### Finding \( \cos(\beta) \): Using the Pythagorean identity: \[ \sin^2(\beta) + \cos^2(\beta) = 1 \] \[ \left(\frac{-2}{5}\right)^2 + \cos^2(\beta) = 1 \] \[ \frac{4}{25} + \cos^2(\beta) = 1 \] \[ \cos^2(\beta) = 1 - \frac{4}{25} = \frac{21}{25} \] \[ \cos(\beta) = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5} \quad (\text{since } \beta \text{ is in quadrant III}) \] ### Step 2: Calculate \( \cos(\alpha+\beta) \) and \( \sin(\alpha+\beta) \) #### Calculate \( \cos(\alpha+\beta) \): \[ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \] Substituting the values: \[ \cos(\alpha+\beta) = \left(\frac{24}{25}\right)\left(-\frac{\sqrt{21}}{5}\right) - \left(-\frac{7}{25}\right)\left(-\frac{2}{5}\right) \] \[ = -\frac{24\sqrt{21}}{125} - \frac{14}{125} \] \[ = \frac{-24\sqrt{21} - 14}{125} \] #### Calculate \( \sin(\alpha+\beta) \): \[ \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \] Substituting the values: \[ \sin(\alpha+\beta) = \left(-\frac{7}{25}\right)\left(-\frac{\sqrt{21}}{5}\right) + \left(\frac{24}{25}\right)\left(-\frac{2}{5}\right) \] \[ = \frac{7\sqrt{21}}{125} - \frac{48}{125} \] \[ = \frac{7\sqrt{21} - 48}{125} \] ### Step 3: Calculate \( \tan(\alpha+\beta) \) Using the values of \( \sin(\alpha+\beta) \) and \( \cos(\alpha+\beta) \): \[ \tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\frac{7\sqrt{21} - 48}{125}}{\frac{-24\sqrt{21} - 14}{125}} = \frac{7\sqrt{21} - 48}{-24\sqrt{21} - 14} \] ### Final Answers: a. \( \cos(\alpha+\beta) = \frac{-24\sqrt{21} - 14}{125} \) b. \( \sin(\alpha+\beta) = \frac{7\sqrt{21} - 48}{125} \) c. \( \tan(\alpha+\beta) = \frac{7\sqrt{21} - 48}{-24\sqrt{21} - 14} \)