A) \[\frac{{{a}^{2}}}{{{b}^{2}}}\]
B) \[\frac{{{b}^{2}}}{{{a}^{2}}}\]
C) \[\frac{a}{b}\]
D) \[\frac{2(2{{a}^{2}}-{{b}^{2}})}{{{b}^{2}}}\]
Correct Answer: D
Solution :
We have, \[\frac{a+\sqrt{{{a}^{2}}-{{b}^{2}}}}{a-\sqrt{{{a}^{2}}-{{b}^{2}}}}+\frac{a-\sqrt{{{a}^{2}}-{{b}^{2}}}}{a+\sqrt{{{a}^{2}}-{{b}^{2}}}}\] Rationalizing both terms, we get \[\frac{{{\left( a+\sqrt{{{a}^{2}}-{{b}^{2}}} \right)}^{2}}}{{{a}^{2}}-{{a}^{2}}+{{b}^{2}}}+\frac{{{\left( a-\sqrt{{{a}^{2}}-{{b}^{2}}} \right)}^{2}}}{{{a}^{2}}-{{a}^{2}}+{{b}^{2}}}\] \[=\frac{1}{{{b}^{2}}}\left[ \begin{align} & {{a}^{2}}+{{a}^{2}}-{{b}^{2}}+2a\sqrt{{{a}^{2}}-{{b}^{2}}}+{{a}^{2}}+{{a}^{2}} \\ & -{{b}^{2}}-2a\sqrt{{{a}^{2}}-{{b}^{2}}} \\ \end{align} \right]\] \[=\frac{2}{{{b}^{2}}}(2{{a}^{2}}-{{b}^{2}})\]
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