`tan(u - v)` Find the exact value of the trigonometric expression given that sin(u) = -7/25 and cos(v) = -4/5 (Both u and v are in quadrant III.)

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`

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`