A) \[1-n\]
B) \[1+n\]
C) \[1-{{n}^{2}}\]
D) \[1+{{n}^{2}}\]
Correct Answer: C
Solution :
\[({{m}^{2}}-{{n}^{2}}){{\sin }^{2}}\beta =\left( \frac{{{\sin }^{2}}\alpha }{{{\sin }^{2}}\beta }-\frac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta } \right){{\sin }^{2}}\beta \] \[=\frac{{{\sin }^{2}}\alpha {{\cos }^{2}}\beta -{{\cos }^{2}}\alpha {{\sin }^{2}}\beta }{{{\sin }^{2}}\beta {{\cos }^{2}}\beta }{{\sin }^{2}}\beta \] \[\because \] \[=\frac{(1-{{\cos }^{2}}\alpha )\,(1-{{\sin }^{2}}\beta )-{{\cos }^{2}}\alpha {{\sin }^{2}}\beta }{{{\cos }^{2}}\beta }\] \[=\frac{1-{{\cos }^{2}}\alpha -{{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\sin }^{2}}\beta -{{\cos }^{2}}\alpha {{\sin }^{2}}\beta }{{{\cos }^{2}}\beta }\] \[=\frac{\cos \beta -{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }=1-\frac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }=1-{{n}^{2}}\]
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