A) \[\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\]
B) \[\frac{m-1}{n-1}\]
C) \[\frac{{{m}^{2}}+1}{{{n}^{2}}+1}\]
D) \[\frac{m+1}{n+1}\]
Correct Answer: A
Solution :
tan A = n tan B \[\Rightarrow \] tan B = \[\frac{1}{n}\] tan A \[\Rightarrow \]cot B =\[\frac{n}{\tan \,A}\]and sin A = m sin B \[\Rightarrow \] sin B=\[\frac{1}{m}\] sin A \[\Rightarrow \]cosec B =\[\frac{m}{\sin \,\,\,A}\] \[\because \]cosec2B \[-\] cot2B =1 \[\Rightarrow \]\[\frac{{{m}^{2}}}{{{\sin }^{2}}A}-\frac{{{n}^{2}}}{{{\tan }^{2}}A}=1\] \[\Rightarrow \]\[\frac{{{m}^{2}}}{{{\sin }^{2}}A}-\frac{{{n}^{2}}{{\cos }^{2}}A}{{{\sin }^{2}}A}=1\] \[\Rightarrow \]\[\frac{{{m}^{2}}-{{n}^{2}}{{\cos }^{2}}A}{{{\sin }^{2}}A}=1\] \[\Rightarrow \]\[{{m}^{2}}-{{n}^{2}}-{{\cos }^{2}}A-{{\cos }^{2}}A\] \[=1-{{\cos }^{2}}A\] \[\Rightarrow \]\[{{m}^{2}}-1={{n}^{2}}{{\cos }^{2}}A-{{\cos }^{2}}A\] \[=({{n}^{2}}-1){{\cos }^{2}}A\] \[\Rightarrow \]\[{{\cos }^{2}}A=\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\]
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