Recall that
[tex]\cot(u+v)=\dfrac{\cos(u+v)}{\sin(u+v)}=\dfrac{\cos u\cos v-\sin u\sin v}{\sin u\cos v+\cos u\sin v}=\dfrac{\cot u\cot v-1}{\cot v+\cot u}[/tex]
Also, recall that for all [tex]\theta[/tex],
[tex]\cos^2\theta+\sin^2\theta=1[/tex]
With [tex]\frac{3\pi}2<u<2\pi[/tex], we can expect [tex]\cos u>0[/tex], and with [tex]0<v<\frac\pi2[/tex], [tex]\sin v>0[/tex]. So from the above identity it follows that
[tex]\cos u=\sqrt{1-\sin^2u}=\dfrac45\implies\cot u=\dfrac{\cos u}{\sin u}=-\dfrac43[/tex]
and
[tex]\sin v=\sqrt{1-\cos^2v}=\dfrac8{17}\implies\cot v=\dfrac{\cos v}{\sin v}=\dfrac{15}8[/tex]
and so
[tex]\cot(u+v)=\dfrac{-\frac43\frac{15}8-1}{\frac{15}8-\frac43}=\boxed{-\dfrac{84}{13}}[/tex]
