Prove that:

$\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1$

To prove: $\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1$

Taking LHS:

$\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}\\[1em] = \dfrac{1}{1+\dfrac{x^a}{x^b}}+\dfrac{1}{1+\dfrac{x^b}{x^a}}\\[1em] = \dfrac{{x^b}}{{x^b}+{x^a}}+\dfrac{{x^a}}{{x^a}+{x^b}}\\[1em] = \dfrac{{x^b}+{x^a}}{{x^b}+{x^a}}\\[1em] = 1 \\[1em] = \text{RHS}$

∴ LHS = RHS

$\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1$