What role does Cauchy's determinant identity play in combinatorics?

Question

When studying representation theory, special functions or various other topics one is very likely to encounter the following identity at some point: $$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}$$ This goes under the name of Cauchy's determinant identity and has various generalizations and analogous statements. There is also a lot of different proofs using either analysis or algebra. In my case I have always seen it introduced (or motivated) as an identity that plays an important role in combinatorics, but I realized that I haven't really seen this identity in a combinatorial context before. In this question I'm asking for a combinatorial interpretation of the above identity. A bonus to someone who can give such an interpretation to Borchardt's variation: $$\det \left(\frac{1}{(x _i+y _j)^2}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)} \cdot \text{per}\left(\frac{1}{x _i +y _j}\right) _{1 \le i,j \le n}$$ (This seems a little too ambitious though, and I would be happy to accept an answer of just the first question)

Answer 1

See also pp. 397--398 of Enumerative Combinatorics, vol. 2. Cauchy's determinant is given in a slightly different but equivalent form. It is explained there that the evaluation of the determinant is equivalent to the fundamental identity $\prod(1-x_iy_j)^{-1} =\sum_\lambda s_\lambda(x)s_\lambda(y)$ in the theory of symmetric functions.

Answer 2

See Theorem 1.5 and section 2 of A Bijective Proof of Borchardt's Identity by Dan Singer.