Question

Let $C$ be right angle of a triangle $ ABC, $ then $ \frac{{{\sin }^{2}}A}{{{\sin }^{2}}B}-\frac{{{\cos }^{2}}A}{{{\cos }^{2}}B} $ is equal to

• A. $ \frac{{{a}^{2}}-{{b}^{2}}}{ab} $
• B. $ \frac{{{a}^{4}}-{{b}^{4}}}{{{a}^{2}}{{b}^{2}}} $
• C. $ \frac{{{a}^{4}}+{{b}^{4}}}{{{a}^{2}}{{b}^{2}}} $
• D. $ \frac{{{a}^{2}}+{{b}^{2}}}{ab} $

Answer 1

Answer: (C)

Given, $ \angle C={{90}^{o}} $
in $ \Delta \,ABC $
$ \because $ $ \cos \,C=\frac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2ab} $
$ \because $ $ \cos \,{{90}^{o}}=\frac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2ab}=0 $
$ \Rightarrow $ $ {{b}^{2}}+{{a}^{2}}-{{c}^{2}}=0\,\,\Rightarrow \,\,\,{{a}^{2}}+{{b}^{2}}={{c}^{2}} $ ..(i)
Now, $ \frac{{{\sin }^{2}}A}{{{\sin }^{2}}B}-\frac{{{\cos }^{2}}A}{{{\cos }^{2}}B} $
$ =\frac{{{(ak)}^{2}}}{{{(bk)}^{2}}}-\frac{{{\left( \frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \right)}^{2}}}{{{\left( \frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} \right)}^{2}}} $
$ =\frac{{{a}^{2}}}{{{b}^{2}}}-\frac{{{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}^{2}}}{{{({{a}^{2}}+{{c}^{2}}-{{b}^{2}})}^{2}}}\times \frac{4\,\,{{a}^{2}}\,{{c}^{2}}}{4\,{{b}^{2}}\,{{c}^{2}}} $
$ =\frac{{{a}^{2}}}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}.{{\left( \frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}} \right)}^{2}} $
$ =\frac{{{a}^{2}}}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}\,{{\left\{ \frac{{{b}^{2}}+{{a}^{2}}+{{b}^{2}}-{{a}^{2}}}{{{a}^{2}}+{{a}^{2}}+{{b}^{2}}-{{b}^{2}}} \right\}}^{2}} $ [from Eq. (i)]
$ =\frac{{{a}^{2}}}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}\left\{ \frac{2{{b}^{2}}}{2{{a}^{2}}} \right\}=\frac{{{a}^{2}}}{{{b}^{2}}}\left\{ 1-\frac{{{b}^{4}}}{{{a}^{4}}} \right\} $
$ =\frac{{{a}^{2}}}{{{b}^{2}}}\times \frac{{{a}^{4}}-{{b}^{4}}}{{{a}^{4}}}=\frac{({{a}^{4}}-{{b}^{4}})}{{{a}^{2}}\,{{b}^{2}}} $