| If \[\frac{\cos \,x}{\cos y}=n\] and \[\frac{\sin \,x}{\sin y}=m,\] then what is the value of \[({{m}^{2}}-{{n}^{2}}){{\sin }^{2}}y\] |
A) \[1-{{n}^{2}}\]
B) \[1+{{n}^{2}}\]
C) \[{{m}^{2}}\]
D) \[{{n}^{2}}\]
Correct Answer: A
Solution :
| \[({{m}^{2}}-{{n}^{2}}){{\sin }^{2}}y=\left( \frac{{{\sin }^{2}}x}{{{\sin }^{2}}y}-\frac{{{\cos }^{2}}x}{{{\cos }^{2}}y} \right){{\sin }^{2}}y\] |
| \[=\,\,\frac{{{\sin }^{2}}x{{\cos }^{2}}y-{{\cos }^{2}}x{{\sin }^{2}}y}{{{\sin }^{2}}y{{\cos }^{2}}y}\times {{\sin }^{2}}y\] |
| \[=\,\,\,\frac{{{\sin }^{2}}x\,(1-{{\sin }^{2}})-(1-{{\sin }^{2}}x)\times {{\sin }^{2}}y}{{{\cos }^{2}}y}\] |
| \[=\,\,\,\frac{{{\sin }^{2}}x-{{\sin }^{2}}x{{\sin }^{2}}y-{{\sin }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y}{{{\cos }^{2}}y}\] |
| \[=\,\,\frac{1-{{\cos }^{2}}x-1+{{\cos }^{2}}y}{{{\cos }^{2}}y}\] |
| \[=\frac{{{\cos }^{2}}y-{{\cos }^{2}}x}{{{\cos }^{2}}y}=1-\frac{{{\cos }^{2}}x}{{{\cos }^{2}}y}=1-{{n}^{2}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
