Question

prove that :

Please do it fast urgent
very urgent!!!!!!​

Answer 1

GIVEN :-

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

TO FIND :-

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

SOLUTION :-

taking LHS

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

know we know that

$\implies \rm{ \bf{x}^{a - b} = \dfrac{ {x}^{a} }{ {x}^{b} } }$

similarly

$\implies \rm{ \bf{x}^{b - a} = \dfrac{ {x}^{b} }{ {x}^{a} } }$

so putting these values :-

$\implies \rm{ \dfrac{1}{1 + \frac{ {x}^{a} }{ {x}^{b} } } + \dfrac{1}{1 + \frac{ {x}^{b} }{ {x}^{a} } } }$

$\implies \rm{ \dfrac{1}{ \frac{ {x}^{b} + {x}^{a} }{ {x}^{b} } } + \dfrac{1}{ \frac{ {x}^{a } + {x}^{b} }{ {x}^{a} } } }$

$\implies \rm{ \dfrac{ {x}^{b} }{ { {x}^{b} + {x}^{a} } } + \dfrac{ {x}^{a} }{ {x}^{b} + {x}^{a} } }$

$\implies \rm{ \dfrac{ {x}^{b} + {x}^{a} }{ { {x}^{b} + {x}^{a} } } }$

$\implies \rm{1 }$

HENCE LHS = RHS

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