1. **State the problem:** We need to find the exact value of $\cos(\alpha + \beta)$ given that $\tan \alpha = \frac{12}{5}$ with $\alpha$ in quadrant III, and $\cos \beta = \frac{15}{17}$ with $\beta$ in quadrant IV. 2. **Recall the formula:** The cosine of a sum is given by $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ 3. **Find $\cos \alpha$ and $\sin \alpha$:** Since $\tan \alpha = \frac{12}{5}$ and $\alpha$ is in quadrant III, both sine and cosine are negative there. - Use the identity $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. - Let $\cos \alpha = x$, then $\sin \alpha = 12k$, $\cos \alpha = 5k$ for some $k$. - Using Pythagoras: $\sin^2 \alpha + \cos^2 \alpha = 1$ gives $$ (12k)^2 + (5k)^2 = 1 \Rightarrow 144k^2 + 25k^2 = 1 \Rightarrow 169k^2 = 1 \Rightarrow k = \pm \frac{1}{13} $$ - Since $\alpha$ is in quadrant III, $\cos \alpha < 0$ and $\sin \alpha < 0$, so $k = -\frac{1}{13}$. - Therefore, $$ \cos \alpha = 5k = 5 \times -\frac{1}{13} = -\frac{5}{13} $$ $$ \sin \alpha = 12k = 12 \times -\frac{1}{13} = -\frac{12}{13} $$ 4. **Find $\sin \beta$:** Given $\cos \beta = \frac{15}{17}$ and $\beta$ in quadrant IV, cosine is positive and sine is negative. - Use Pythagoras: $\sin^2 \beta + \cos^2 \beta = 1$. - So, $$ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289} $$ - Since $\beta$ is in quadrant IV, $\sin \beta < 0$, so $$ \sin \beta = -\frac{8}{17} $$ 5. **Calculate $\cos(\alpha + \beta)$:** $$ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \left(-\frac{5}{13}\right) \left(\frac{15}{17}\right) - \left(-\frac{12}{13}\right) \left(-\frac{8}{17}\right) $$ 6. **Simplify:** $$ = -\frac{75}{221} - \frac{96}{221} = -\frac{75 + 96}{221} = -\frac{171}{221} $$ 7. **Final answer:** $$ \boxed{-\frac{171}{221}} $$