A) $ \frac{m^{2}}{n^{2}} $
B) $ \frac{m^{2}-1}{n^{2}-1} $
C) $ \frac{m^{2}+1}{n^{2}+1} $
D) $ \frac{m^{2}}{n^{2}+1} $
Correct Answer: B
$ \Rightarrow $ $ \frac{\sin \alpha }{\cos \alpha }=n\frac{\sin \beta }{\cos \beta } $
$ \Rightarrow $ $ \frac{m\sin \beta }{\cos \alpha }=n\frac{\sin \beta }{\cos \beta } $ $ [\because \sin \alpha =m\sin \beta ] $
$ \Rightarrow $ $ \cos \alpha =\frac{m}{n}\cos \beta $ On squaring both sides, we get $ {{\cos }^{2}}\alpha =\frac{m^{2}}{n^{2}}{{\cos }^{2}}\beta $ (i) Also, $ \sin \alpha =m\sin \beta $ On squaring both sides, we get $ {{\sin }^{2}}\alpha =m^{2}{{\sin }^{2}}\beta $
$ \Rightarrow $ $ 1-{{\cos }^{2}}\alpha =m^{2}(1-{{\cos }^{2}}\beta ) $
$ \Rightarrow $ $ 1-{{\cos }^{2}}\alpha =m^{2}-m^{2}{{\cos }^{2}}\beta $
$ \Rightarrow $ $ -\frac{(1-{{\cos }^{2}}\alpha -m^{2})}{m^{2}}={{\cos }^{2}}\beta $
$ \Rightarrow $ $ \frac{({{\cos }^{2}}\alpha +m^{2}-1)}{m^{2}}={{\cos }^{2}}\beta $ (ii) From Eqs. (i) and (ii), we get $ {{\cos }^{2}}\alpha =\frac{m^{2}}{n^{2}}\times \frac{({{\cos }^{2}}\alpha +m^{2}-1)}{m^{2}} $
$ \Rightarrow $ $ n^{2}{{\cos }^{2}}\alpha ={{\cos }^{2}}\alpha +m^{2}-1 $
$ \Rightarrow $ $ (n^{2}-1){{\cos }^{2}}\alpha =m^{2}-1 $
$ \therefore $ $ {{\cos }^{2}}\alpha =\frac{m^{2}-1}{n^{2}-1} $
BETA
