Question

$\frac{1}{1 + {x}^{a - b} } + \frac{1}{1 + {x}^{b - a} }$
is equal to ?

Answer 1

$\frac{1}{1+x^{a-b}}+\frac{1}{1 + x^{b-a}} \\ \\ = \frac{1}{1 + {x}^{a - b} } + \frac{1}{1 + {x}^{ - (a - b)} } \\ \\ = \frac{1}{1 + {x}^{a - b} } + \frac{1}{1 + \frac{1}{ {x}^{a - b} } } \\ \\ = \frac{1}{1 + {x}^{a - b} } + \frac{ {x}^{a - b} }{ {x}^{a - b} + 1 } \\ \\ = \frac{(1 + {x}^{a - b}) }{1 + {x}^{a - b} } \\ \\ = 1$

Answer 2

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

$= \frac{1}{1 + {x}^{- (b - a) } } + \frac{1}{1 + {x}^{b - a} }$

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

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

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

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

$= 1$