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) \[\frac{{{a}^{2}}+{{b}^{2}}}{ab}\]
Correct Answer: C
Solution :
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)] \[=\frac{{{a}^{2}}}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}\left\{ \frac{2{{b}^{2}}}{2{{a}^{2}}} \right\}=\frac{{{a}^{2}}}{{{b}^{2}}}\left\{ 1-\frac{{{b}^{4}}}{{{a}^{4}}} \right\}\] \[=\frac{{{a}^{2}}}{{{b}^{2}}}\times \frac{{{a}^{4}}-{{b}^{4}}}{{{a}^{4}}}=\frac{({{a}^{4}}-{{b}^{4}})}{{{a}^{2}}\,{{b}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
