1. **Problem statement:** Find the exact value of $ \cos(u - v) $ given $ \csc u = -\frac{13}{12} $ with $ \pi < u < \frac{3\pi}{2} $, and $ \cos v = \frac{4}{5} $ where $ v $ is in Quadrant I. 2. **Recall formulas and quadrant rules:** - $ \csc u = \frac{1}{\sin u} $ so $ \sin u = \frac{1}{\csc u} = -\frac{12}{13} $. - Since $ u $ is in Quadrant III (between $ \pi $ and $ \frac{3\pi}{2} $), both sine and cosine are negative. - Use Pythagorean identity: $ \sin^2 u + \cos^2 u = 1 $ to find $ \cos u $. - $ v $ is in Quadrant I, so $ \sin v > 0 $ and $ \cos v > 0 $. - Use $ \sin^2 v + \cos^2 v = 1 $ to find $ \sin v $. - Use the cosine difference formula: $ \cos(u - v) = \cos u \cos v + \sin u \sin v $. 3. **Calculate $ \cos u $:** $$ \cos^2 u = 1 - \sin^2 u = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} $$ Since $ u $ is in Quadrant III, $ \cos u < 0 $, so $$ \cos u = -\frac{5}{13} $$ 4. **Calculate $ \sin v $:** $$ \sin^2 v = 1 - \cos^2 v = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} $$ Since $ v $ is in Quadrant I, $ \sin v > 0 $, so $$ \sin v = \frac{3}{5} $$ 5. **Apply cosine difference formula:** $$ \cos(u - v) = \cos u \cos v + \sin u \sin v = \left(-\frac{5}{13}\right)\left(\frac{4}{5}\right) + \left(-\frac{12}{13}\right)\left(\frac{3}{5}\right) $$ Simplify step-by-step: $$ = -\frac{20}{65} - \frac{36}{65} = -\frac{56}{65} $$ **Final answer:** $$ \boxed{-\frac{56}{65}} $$