Assumint that u is in the 1st quadrant and v is in the second quadrant, then given that
[tex]\sin u= \frac{8}{17} \\ \\ \Rightarrow\cos u= \frac{ \sqrt{17^2-8^2} }{17} \\ \\ = \frac{ \sqrt{289-64} }{17} = \frac{ \sqrt{225} }{17} = \frac{15}{17} [/tex]
Given that
[tex]\cos v=- \frac{60}{61} \\ \\ \Rightarrow\sin v= \frac{ \sqrt{61^2-60^2} }{61} \\ \\ = \frac{ \sqrt{3,721-3,600} }{61} = \frac{ \sqrt{121} }{61} \\ \\ = \frac{11}{61} [/tex]
[tex]\cos(u+v)=\cos u\cos v-\sin u\sin v \\ \\ = \frac{15}{17} \times \left(-\frac{60}{61}\right) - \frac{8}{17} \times \frac{11}{61} \\ \\ =- \frac{900}{1,037} - \frac{88}{1,037} = -\frac{988}{1,037} [/tex]
