Answer
The identity is verified.
$\frac{cos~x-cos~y}{sin~x+sin~y}+\frac{sin~x-sin~y}{cos~x+cos~y}=0$
Work Step by Step
$\frac{cos~x-cos~y}{sin~x+sin~y}+\frac{sin~x-sin~y}{cos~x+cos~y}=\frac{cos~x-cos~y}{sin~x+sin~y}~\frac{cos~x+cos~y}{cos~x+cos~y}+\frac{sin~x-sin~y}{cos~x+cos~y}~\frac{sin~x+sin~y}{sin~x+sin~y}=\frac{cos^2x-cos^2y+sin^2x-sin^2y}{(cos~x+cos~y)(sin~x+sin~y)}=\frac{(cos^2x+sin^2x)-(sin^2y+cos^2y)}{(cos~x+cos~y)(sin~x+sin~y)}=\frac{1-1}{(cos~x+cos~y)(sin~x+sin~y)}=0$
