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If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{H-a}+\frac{1}{H-b}$ is
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Verified Answer
The correct answer is:
$\frac{1}{a}+\frac{1}{b}$
Putting $H=\frac{2 a b}{a+b}$, we have $\frac{1}{H-a}+\frac{1}{H-b}$
$=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}$ $=\frac{a+b}{a b-a^2}+\frac{a+b}{a b-b^2}$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$
$=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}$ $=\frac{a+b}{a b-a^2}+\frac{a+b}{a b-b^2}$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$
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