Answer
$sin~(x+y) = \frac{4-9\sqrt{11}}{50}$
$cos~(x-y) = \frac{12\sqrt{11}-3}{50}$
$tan~(x+y) = \frac{4-9\sqrt{11}}{12\sqrt{11}+3}$
$x+y~~$ is in quadrant IV.
Work Step by Step
$sin~x = \frac{1}{10}$
Since $x$ is in quadrant I, $cos~x$ is also positive.
We can find $cos~x$:
$sin^2~x+cos^2~x = 1$
$cos~x = \sqrt{1-sin^2~x}$
$cos~x = \sqrt{1-(\frac{1}{10})^2}$
$cos~x = \sqrt{1-\frac{1}{100}}$
$cos~x = \frac{3\sqrt{11}}{10}$
$cos~y = \frac{4}{5}$
Since $y$ is in quadrant IV, $sin~y$ is negative.
We can find $sin~y$:
$sin^2~y+cos^2~y = 1$
$sin~y = -\sqrt{1-cos^2~y}$
$sin~y = -\sqrt{1-(\frac{4}{5})^2}$
$sin~y = -\sqrt{1-\frac{16}{25}}$
$sin~y = -\frac{3}{5}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (\frac{1}{10})(\frac{4}{5})+(\frac{3\sqrt{11}}{10})(-\frac{3}{5})$
$sin~(x+y) = \frac{2}{25}-\frac{9\sqrt{11}}{50}$
$sin~(x+y) = \frac{4-9\sqrt{11}}{50}$
We can find $cos(x-y)$:
$cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$
$cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$
$cos~(x-y) = (\frac{3\sqrt{11}}{10})(\frac{4}{5})-(\frac{1}{10})(-\frac{-3}{5})$
$cos~(x-y) = \frac{12\sqrt{11}}{50}-\frac{3}{50}$
$cos~(x-y) = \frac{12\sqrt{11}-3}{50}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (\frac{3\sqrt{11}}{10})(\frac{4}{5})-(\frac{1}{10})(-\frac{3}{5})$
$cos~(x+y) = \frac{12\sqrt{11}}{50}+\frac{3}{50}$
$cos~(x+y) = \frac{12\sqrt{11}+3}{50}$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{\frac{4-9\sqrt{11}}{50}}{\frac{12\sqrt{11}+3}{50}}$
$tan~(x+y) = \frac{4-9\sqrt{11}}{12\sqrt{11}+3}$
Since $sin~(x+y)$ and $tan~(x+y)$ are both negative, then $x+y$ must be in quadrant IV.