A) \[\frac{{{a}^{2}}-1}{{{b}^{2}}-1}\]
B) \[\frac{{{a}^{2}}+1}{{{b}^{2}}-1}\]
C) \[\frac{{{b}^{2}}-1}{{{a}^{2}}+1}\]
D) \[\frac{{{b}^{2}}-1}{{{a}^{2}}-1}\]
Correct Answer: D
Solution :
Given, \[\tan \,A=a\,\tan B\] and \[\sin A=b\,\sin B\] \[\cot \,B=\frac{a}{\tan A}\] and \[\cos ecB=\frac{b}{\sin A}\] We know that, \[\cos e{{c}^{2}}B-{{\cot }^{2}}B=1\] \[{{\left( \frac{b}{\sin A} \right)}^{2}}-{{\left( \frac{a}{\tan A} \right)}^{2}}=1\] \[\frac{{{b}^{2}}}{{{\sin }^{2}}A}-\frac{{{a}^{2}}}{{{\tan }^{2}}A}=1\] \[\frac{1}{{{\sin }^{2}}A}\left[ {{b}^{2}}-{{a}^{2}}{{\cos }^{2}}A \right]=1\] \[{{b}^{2}}-{{a}^{2}}{{\cos }^{2}}A={{\sin }^{2}}A\] \[{{b}^{2}}=1-{{\cos }^{2}}A+{{a}^{2}}{{\cos }^{2}}A\] \[{{\cos }^{2}}A\left( {{a}^{2}}-1 \right)=\left( {{b}^{2}}-1 \right)\] \[\Leftrightarrow \,\,{{\cos }^{2}}A=\frac{{{b}^{2}}-1}{{{a}^{2}}-1}\]
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