A) \[a+b\]
B) \[ab\]
C) \[\frac{1}{a}+\frac{1}{b}\]
D) \[\frac{1}{a}-\frac{1}{b}\]
Correct Answer: C
Solution :
Putting\[H=\frac{2ab}{a+b}\], we have \[\frac{1}{H-a}+\frac{1}{H-b}\] \[=\frac{1}{\left( \frac{2ab}{a+b}-a \right)}+\frac{1}{\left( \frac{2ab}{a+b}-b \right)}=\frac{a+b}{ab-{{a}^{2}}}+\frac{a+b}{ab-{{b}^{2}}}\] \[=\left( \frac{a+b}{b-a} \right)\left( \frac{1}{a}-\frac{1}{b} \right)=\left( \frac{a+b}{b-a} \right)\left( \frac{b-a}{ab} \right)=\frac{a+b}{ab}=\frac{1}{a}+\frac{1}{b}\].
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