| If \[\tan \alpha =n\tan \beta \]and \[\sin \alpha =m\sin \beta ,\]then \[{{\cos }^{2}}\alpha \]is [SSC (CGL) 2013] |
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
Solution :
| \[\tan \alpha =n\tan \beta \] |
| \[\Rightarrow \] \[\frac{\sin \alpha }{\cos \alpha }=n\frac{\sin \beta }{\cos \beta }\] |
| \[\Rightarrow \] \[\frac{m\sin \beta }{\cos \alpha }=n\frac{\sin \beta }{\cos \beta }\]\[[\because \sin \alpha =m\sin \beta ]\] |
| \[\Rightarrow \] \[\cos \alpha =\frac{m}{n}\cos \beta \] |
| 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 |
| \[{{\cos }^{2}}\alpha =\frac{{{m}^{2}}}{{{n}^{2}}}\times \frac{({{\cos }^{2}}\alpha +{{m}^{2}}-1)}{{{m}^{2}}}\] |
| \[\Rightarrow \] \[{{n}^{2}}{{\cos }^{2}}\alpha ={{\cos }^{2}}\alpha +{{m}^{2}}-1\] |
| \[\Rightarrow \]\[({{n}^{2}}-1){{\cos }^{2}}\alpha ={{m}^{2}}-1\] |
| \[\therefore \] \[{{\cos }^{2}}\alpha =\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\] |
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