Question

plz solve this question plz...

Answer 1

$\pink{\huge{\underline{\star\:Answer\: \star}}}$

$\frac{1}{1+x^{(a-b)}}+\frac{1}{1+x^{(b-a)}}=1\\\implies Take\: L.H.S\colon \\\\\implies \frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$

$\\\implies \frac{1}{\frac{x^{b}+x^{a}}{x^{b}}}+\frac{1}{\frac{x^{a}+x^{b}}{x^{a}}}\\\implies \frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}\\\implies \frac{\cancel{(x^{a}+x^{b})}}{\cancel{(x^{a}+x^{b})}}\\\implies 1$

$\red{\underline{Answer}}$

$\pink{\boxed{{L.H.S=R.H.S.}}}$

$\green{\bold{{Hence \:Proved}}}$

Answer 2

Hello!

$\bf{(R.H.S.)}$

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

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

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

= $\dfrac{\cancel{(x^{a}+x^{b})}}{\cancel{(x^{a}+x^{b})}}$ = 1

$\bf{(L.H.S.)}$

$\boxed{\red{\bf{HENCE\:PROVED}}}$

Cheers!