A) \[\frac{c+d}{cd}\]
B) \[\frac{cd}{c+d}\]
C) \[\frac{c-d}{c+d}\]
D) \[\frac{c+d}{c-d}\]
Correct Answer: D
Solution :
\[\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}=\frac{ab}{cd}\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}}{ab}=\frac{{{c}^{2}}+{{d}^{2}}}{cd}\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}}{ab}=\frac{{{c}^{2}}+{{d}^{2}}}{2cd}\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}+2ac}{{{a}^{2}}+{{b}^{2}}-2ab}=\frac{{{c}^{2}}+{{d}^{2}}+2cd}{{{c}^{2}}+{{d}^{2}}-2cd}\] (by componendo and dividendo) \[\Rightarrow \] \[\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{{{(c+d)}^{2}}}{{{(c-d)}^{2}}}\] \[\therefore \] \[\frac{a+b}{a-b}=\frac{c+d}{c-d}\]
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