Precalculus, Chapter 5, 5.4, Section 5.4, Problem 49

Given sin(u)=-7/25 , cos(v)=-4/5
using pythagorean identity,
sin^2(u)+cos^2(u)=1
(-7/25)^2+cos^2(u)=1
cos^2(u)=1-49/625=(625-49)/625=576/625
cos(u)=sqrt(576/625)
cos(u)=+-24/25
since u is in quadrant III,
:.cos(u)=-24/25
sin^2(v)+cos^2(v)=1
sin^2(v)+(-4/5)^2=1
sin^2(v)+16/25=1
sin^2(v)=1-16/25=(25-16)/25=9/25
sin(v)=sqrt(9/25)
sin(v)=+-3/5
since v is in quadrant III,
:.sin(v)=-3/5
Now let's evaluate tan(u-v),
tan(u-v)=sin(u-v)/cos(u-v)
=(sin(u)cos(v)-cos(u)sin(v))/(cos(u)cos(v)+sin(u)sin(v))
=((-7/25)(-4/5)-(-24/25)(-3/5))/((-24/25)(-4/5)+(-7/25)(-3/5))
=(28/125-72/125)/(96/125+21/125)
=(-44/125)/(117/125)
=-44/117