A) 0
B) 1
C) \[\frac{2{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}\]
D) \[\frac{2{{b}^{2}}}{{{b}^{2}}+{{a}^{2}}}\]
Correct Answer: C
Solution :
(c): We have, \[\frac{{{a}^{-1}}}{{{a}^{-1}}+{{b}^{-1}}}+\frac{{{a}^{-1}}}{{{a}^{-1}}-{{b}^{-1}}}\] \[=\frac{1/a}{\frac{1}{a}+\frac{1}{b}}+\frac{1/a}{\frac{1}{a}-\frac{1}{b}}=\frac{1/a}{\frac{b+a}{ab}}+\frac{1/a}{\frac{b-a}{ab}}\] \[=\frac{1}{a}.\frac{ab}{b+a}+\frac{1}{a}.\frac{ab}{b-a}=\frac{b}{b+a}+\frac{b}{b-a}\] \[=\frac{b(b-a)+b(b+a)}{(b+a)(b-a)}=\frac{{{b}^{2}}-ab+{{b}^{2}}+ab}{{{b}^{2}}-{{a}^{2}}}\] \[=\frac{2{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}\]
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