Answer
$cos(u-v)=-1$
Work Step by Step
$1+tan^2u=sec^2u$
$1+\frac{9}{16}=sec^2u$
$\frac{25}{16}=\frac{1}{cos^2u}$
$cos^2u=\frac{16}{25}$
$cos~u=±\frac{4}{5}~~$ (But, $u$ is in Quadrant I):
$cos~u=\frac{4}{5}$
$cos^2u+sin^2u=1$
$\frac{16}{25}+sin^2u=1$
$sin^2u=\frac{9}{25}$
$sin~u=±\frac{3}{5}~~$ (But, $u$ is in Quadrant I):
$sin~u=\frac{3}{5}$
$cos^2v+sin^2v=1$
$\frac{16}{25}+sin^2v=1$
$sin^2v=\frac{9}{25}$
$sin~v=±\frac{3}{5}~~$ (But, $v$ is in Quadrant III):
$sin~v=-\frac{3}{5}$
$cos(u-v)=cos~u~cos~v+sin~u~sin~v=\frac{4}{5}(-\frac{4}{5})+\frac{3}{5}(-\frac{3}{5})=-\frac{16}{25}-\frac{9}{25}=-\frac{25}{25}=-1$
