Answer
The identity is verified.
$\frac{tan~x+tan~y}{1-tan~x~tan~y}=\frac{cot~x+cot~y}{cot~x~cot~y-1}$
Work Step by Step
Remember:
$tan~x=\frac{1}{cot~x}$
$tan~x~cot~x=1$
$\frac{tan~x+tan~y}{1-tan~x~tan~y}=\frac{tan~x+tan~y}{1-tan~x~tan~y}~\frac{cot~x~cot~y}{cot~x~cot~y}=\frac{(tan~x~cot~x)~cot~y+cot~x~(tan~y~cot~y)}{cot~x~cot~y-(tan~x~cot~x)(tan~y~cot~y)}=\frac{cot~y+cot~x}{cot~x~cot~y-1}$
