A) \[\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\]
B) \[\frac{{{m}^{2}}+1}{{{n}^{2}}+1}\]
C) \[\frac{{{m}^{2}}-1}{{{n}^{2}}+1}\]
D) \[\frac{{{m}^{2}}+1}{{{n}^{2}}-1}\]
Correct Answer: A
Solution :
| [a] \[\tan A=n\tan \,B\] \[\Rightarrow \] \[\frac{\sin A}{\cos A}=n\frac{\sin B}{\cos B}\]\[\Rightarrow \]\[n=\frac{\sin A\cdot \cos B}{\operatorname{cosA}\cdot \sin B}\] \[\Rightarrow \] \[n=\frac{\sin A}{\sin B}\times \frac{\cos B}{\cos A}\] \[\Rightarrow \] \[n=m\times \frac{\sqrt{1-{{\sin }^{2}}B}}{\cos A}\]\[(\because sin\,A=m\,sin\,B)\] \[\Rightarrow \] \[\cos A\times n=m\times \sqrt{1-{{\sin }^{2}}B}\] \[\Rightarrow \] \[{{n}^{2}}{{\cos }^{2}}A={{m}^{2}}(1-si{{n}^{2}}B)\] \[\Rightarrow \] \[{{n}^{2}}{{\cos }^{2}}A={{m}^{2}}\left( 1-\frac{{{\sin }^{2}}A}{{{m}^{2}}} \right)\] \[\Rightarrow \] \[{{n}^{2}}{{\cos }^{2}}A={{m}^{2}}-{{\sin }^{2}}A\] \[\Rightarrow \] \[{{n}^{2}}{{\cos }^{2}}A-{{\cos }^{2}}A+{{\cos }^{2}}A+{{\sin }^{2}}A={{m}^{2}}\] \[\Rightarrow \] \[({{n}^{2}}-1)co{{s}^{2}}A+1={{m}^{2}}\]\[\Rightarrow \] \[\Rightarrow \] \[({{n}^{2}}-1)co{{s}^{2}}A={{m}^{2}}-1\] \[\therefore \] \[{{\cos }^{2}}A=\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\] |
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