Question

Evaluate the given expression:
$\frac{1}{1 + {a}^{x - y} } + \frac{1}{1 + {a}^{y - x} }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \dfrac{1}{1 + {a}^{x - y} } + \dfrac{1}{1 + {a}^{y - x} }$

We know,

$\color{green}\boxed{ \rm{ \: {a}^{m - n} = \frac{ {a}^{m} }{ {a}^{n} } \: }}$

So, using this identity, the above expression can be rewritten as

$\rm \: = \: \dfrac{1}{1 + \dfrac{ {a}^{x} }{ {a}^{y} } } \: + \: \dfrac{1}{1 + \dfrac{ {a}^{y} }{ {a}^{x} } }$

$\rm \: = \: \dfrac{1}{\dfrac{{a}^{y} + {a}^{x} }{ {a}^{y} } } \: + \: \dfrac{1}{ \dfrac{{a}^{x} + {a}^{y} }{ {a}^{x} } }$

$\rm \: = \: \dfrac{{a}^{y}}{{a}^{y} + {a}^{x}} + \dfrac{{a}^{x}}{{a}^{x} + {a}^{y}}$

can be rewritten as

$\rm \: = \: \dfrac{{a}^{y}}{{a}^{y} + {a}^{x}} + \dfrac{{a}^{x}}{{a}^{y} + {a}^{x}}$

$\rm \: = \: \dfrac{{a}^{y} + {a}^{x}}{{a}^{y} + {a}^{x}}$

$\rm \: = \: 1$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \dfrac{1}{1 + {a}^{x - y} } + \dfrac{1}{1 + {a}^{y - x} } \: = \: 1 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$