Answer
$sin~(x+y) = \frac{44}{125}$
$cos~(x-y) = \frac{3}{5}$
$tan~(x+y) = \frac{44}{117}$
$x+y~~$ is in quadrant I
Work Step by Step
$sin~x = \frac{3}{5}$
Since $x$ is in quadrant I, $cos~x$ is also positive.
We can find $cos~x$:
$sin^2~x+cos^2~x = 1$
$cos~x = \sqrt{1-sin^2~x}$
$cos~x = \sqrt{1-(\frac{3}{5})^2}$
$cos~x = \sqrt{1-\frac{9}{25}}$
$cos~x = \frac{4}{5}$
$cos~y = \frac{24}{25}$
Since $y$ is in quadrant IV, $sin~y$ is negative.
We can find $sin~y$:
$sin^2~y+cos^2~y = 1$
$sin~y = -\sqrt{1-cos^2~y}$
$sin~y = -\sqrt{1-(\frac{24}{25})^2}$
$sin~y = -\sqrt{1-\frac{576}{625}}$
$sin~y = -\frac{7}{25}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (\frac{3}{5})(\frac{24}{25})+(\frac{4}{5})(-\frac{7}{25})$
$sin~(x+y) = \frac{72}{125}-\frac{28}{125}$
$sin~(x+y) = \frac{44}{125}$
We can find $cos(x-y)$:
$cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$
$cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$
$cos~(x-y) = (\frac{4}{5})(\frac{24}{25})-(\frac{3}{5})(-\frac{-7}{25})$
$cos~(x-y) = \frac{96}{125}-\frac{21}{125}$
$cos~(x-y) = \frac{75}{125}$
$cos~(x-y) = \frac{3}{5}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (\frac{4}{5})(\frac{24}{25})-(\frac{3}{5})(\frac{-7}{25})$
$cos~(x+y) = \frac{96}{125}+\frac{21}{125}$
$cos~(x+y) = \frac{117}{125}$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{ \frac{44}{125}}{\frac{117}{125}}$
$tan~(x+y) = \frac{44}{117}$
Since $sin~(x+y)$ and $tan~(x+y)$ are both positive, then $x+y$ must be in quadrant I.
