Question

prove the following equation

Answer 1

Given to prove:-

$\dfrac{a^{-1}}{a^{-1}+b^{-1}} +\dfrac{a^{-1}}{a^{-1}-b^{-1}} = \dfrac{2b^2}{b^2-a^2}$

SOLUTION :-

Take L.H.S

We shall take each term separately and solve it

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

As we know that from exponent and laws

$a^{-n} = \dfrac{1}{a^n}$

So,

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

$=\dfrac{\dfrac{1}{a} }{\dfrac{a+b}{ab} }$

$=\dfrac{ab}{(a)(a+b)}$

$\dfrac{ab}{a^2+ab}$

So,

$\dfrac{a^{-1}}{a^{-1}+b^{-1}}=\dfrac{ab}{a^2+ab}$

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Now we shall solve 2nd term

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

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

$=\dfrac{\dfrac{1}{a} }{\dfrac{b-a}{ab} }$

$= \dfrac{ab}{(a)(b-a)}$

$=\dfrac{ab}{ab-a^2}$

So,

$\dfrac{a^{-1}}{a^{-1}-b^{-1}} =\dfrac{ab}{ab-a^2}$

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Now,

$\dfrac{a^{-1}}{a^{-1}+b^{-1}} +\dfrac{a^{-1}}{a^{-1}-b^{-1}} =\dfrac{ab}{a^2+ab}+\dfrac{ab}{ab-a^2}$

$=\dfrac{ab(ab-a^2)+ab(a^2+ab)}{(ab+a^2)(ab-a^2)}$

Denominator is in form of (a+b)(a-b) = a²-b²

$= \dfrac{a^2b^2 -a^3b + a^3b + a^2b^2}{a^2b^2 -a^4}$

Take common a² to the denominator

$= \dfrac{a^2b^2 + a^2b^2}{a^2(b^2 -a^2)}$

$= \dfrac{2a^2b^2 }{a^2(b^2 -a^2)}$

$= \dfrac{2b^2 }{(b^2 -a^2)}$

L.H.S = R.H.S

Hence proved !