Question

prove it plz plz plz plz​

Answer 1

Solution

$\bf We \:\:have$

$\tt\to \dfrac{ x^{-1}}{ x^{-1}+y^{-1}} + \dfrac{x^{-1}}{ x^{-1}-y^{-1}}=\dfrac{2y^2}{y^2-x^2}$

$\bf Taking\:\:LCM\:\: we\:\: get$

$\tt\to \dfrac{x^{-1}(x^{-1}-y^{-1})+x^{-1}(x^{-1}+y^{-1})}{(x^{-1}+ y^{-1})(x^{-1}-y^{-1})}$

$\bf Using\:\:Some\:\:identities$

$\to\tt (a+b)(a-b)=a^2-b^2$

$\to\tt a(c+b) = ac+ab$

$\bf Now\:\:we\:\: get$

$\tt\to \dfrac{x^{-2}- x^{-1}y^{-1}+x^{-2}+ x^{-1}y^{-1}}{(x^{-2}-y^{-2})}$

$\tt\to\dfrac{2x^{-2}}{x^{-2}-y^{-2}}$

$\tt\to\dfrac{\dfrac{2}{x^2} }{\dfrac{1}{x^2}-\dfrac{1}{y^2} }$

$\to\tt\dfrac{\dfrac{2}{x^2} }{\dfrac{y^2-x^2}{x^2y^2} }$

$\tt\to\dfrac{2}{\dfrac{x^2(y^2-x^2)}{x^2y^2} }$

$\tt\to\dfrac{2}{\dfrac{(y^2-x^2)}{y^2} }$

$\tt\to\dfrac{2y^2}{{(y^2-x^2)}}$

$\bf Hence\:\:Proved$

Answer 2

Answer:

hey mate

Step-by-step explanation:

Solution

\bf We \:\:haveWehave

\tt\to \dfrac{ x^{-1}}{ x^{-1}+y^{-1}} + \dfrac{x^{-1}}{ x^{-1}-y^{-1}}=\dfrac{2y^2}{y^2-x^2}→x−1+y−1x−1+x−1−y−1x−1=y2−x22y2

\bf Taking\:\:LCM\:\: we\:\: getTakingLCMweget

\tt\to \dfrac{x^{-1}(x^{-1}-y^{-1})+x^{-1}(x^{-1}+y^{-1})}{(x^{-1}+ y^{-1})(x^{-1}-y^{-1})}→(x−1+y−1)(x−1−y−1)x−1(x−1−y−1)+x−1(x−1+y−1)

\bf Using\:\:Some\:\:identitiesUsingSomeidentities

\to\tt (a+b)(a-b)=a^2-b^2→(a+b)(a−b)=a2−b2

\to\tt a(c+b) = ac+ab→a(c+b)=ac+ab

\bf Now\:\:we\:\: getNowweget

\tt\to \dfrac{x^{-2}- x^{-1}y^{-1}+x^{-2}+ x^{-1}y^{-1}}{(x^{-2}-y^{-2})}→(x−2−y−2)x−2−x−1y−1+x−2+x−1y−1

\tt\to\dfrac{2x^{-2}}{x^{-2}-y^{-2}}→x−2−y−22x−2

\tt\to\dfrac{\dfrac{2}{x^2} }{\dfrac{1}{x^2}-\dfrac{1}{y^2} }→x21−y21x22

\to\tt\dfrac{\dfrac{2}{x^2} }{\dfrac{y^2-x^2}{x^2y^2} }→x2y2y2−x2x22

\tt\to\dfrac{2}{\dfrac{x^2(y^2-x^2)}{x^2y^2} }→x2y2x2(y2−x2)2

\tt\to\dfrac{2}{\dfrac{(y^2-x^2)}{y^2} }→y2(y2−x2)2

\tt\to\dfrac{2y^2}{{(y^2-x^2)}}→(y2−x2)2y2

\bf Hence\:\:ProvedHenceProved