Question

solve it please as fast as you can
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Answer 1

Consider the term,

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

Well we know that $x^{-1}=\dfrac {1}{x}.$ Thus,

$\dfrac {a^{-1}}{a^{-1}+b^{-1}}=\dfrac {\left (\dfrac {1}{a}\right)}{\left (\dfrac {1}{a}+\dfrac {1}{b}\right)}\\\\\\\dfrac {a^{-1}}{a^{-1}+b^{-1}}=\dfrac {\left (\dfrac {b}{ab}\right)}{\left (\dfrac {a+b}{ab}\right)}\\\\\\\dfrac {a^{-1}}{a^{-1}+b^{-1}}=\dfrac {b}{a+b}$

Similarly,

$\dfrac {a^{-1}}{a^{-1}-b^{-1}}=\dfrac {\left (\dfrac {1}{a}\right)}{\left (\dfrac {1}{a}-\dfrac {1}{b}\right)}\\\\\\\dfrac {a^{-1}}{a^{-1}-b^{-1}}=\dfrac {\left (\dfrac {b}{ab}\right)}{\left (\dfrac {b-a}{ab}\right)}\\\\\\\dfrac {a^{-1}}{a^{-1}-b^{-1}}=\dfrac {b}{b-a}$

Then,

$\begin {aligned}&\textsf {LHS}\\\\\implies\ \ &\dfrac {a^{-1}}{a^{-1}+b^{-1}}+\dfrac {a^{-1}}{a^{-1}-b^{-1}}\\\\\implies\ \ &\dfrac {b}{a+b}+\dfrac {b}{b-a}\\\\\implies\ \ &b\left (\dfrac {1}{a+b}-\dfrac {1}{a-b}\right)\\\\\implies\ \ &b\left (\dfrac {(a-b)-(a+b)}{(a+b)(a-b)}\right)\\\\\implies\ \ &b\left (\dfrac {a-b-a-b}{a^2-b^2}\right)\\\\\implies\ \ &b\left (\dfrac {-2b}{a^2-b^2}\right)\\\\\implies\ \ &\dfrac {-\left(2b^2\right)}{a^2-b^2}\\\\\implies\ \ &\textsf {RHS}\end {aligned}$

Hence Proved!

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