Precalculus, Chapter 5, 5.4, Section 5.4, Problem 46

Given sin(u)=5/13 , cos(v)=-3/5
using pythagorean identity,
sin^2(u)+cos^2(u)=1
(5/13)^2+cos^2(u)=1
cos^2(u)=1-25/169=(169-25)/169=144/169
cos(u)=sqrt(144/169)
cos(u)=+-12/13
since u is in quadrant II,
:.cos(u)=-12/13
sin^2(v)+cos^2(v)=1
sin^2(v)+(-3/5)^2=1
sin^2(v)+9/25=1
sin^2(v)=1-9/25=(25-9)/25=16/25
sin(v)=sqrt(16/25)
sin(v)=+-4/5
since v is in quadrant II,
:.sin(v)=4/5
Now let's evaluate cot(u+v),
cot(u+v)=cos(u+v)/sin(u+v)
=(cos(u)cos(v)-sin(u)sin(v))/(sin(u)cos(v)+cos(u)sin(v))
=((-12/13)(-3/5)-(5/13)(4/5))/((5/13)(-3/5)+(-12/13)(4/5))
=(36/65-20/65)/(-15/65-48/65)
=-16/63