A) \[{{\alpha }^{2}}+{{\beta }^{2}}\]
B) \[{{\alpha }^{2}}-{{\beta }^{2}}\]
C) 1
D) 0
Correct Answer: D
Solution :
On solving the determinant, \[1\,(1-{{\cos }^{2}}\beta )-\cos (\alpha -\beta )\]\[[\cos (\alpha -\beta )-\cos \alpha \cos \beta ]\]\[+\cos \alpha [\cos \beta \cos (\alpha -\beta )-\cos \alpha ]\] \[=1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha -{{\cos }^{2}}(\alpha -\beta )\]\[+2\cos \alpha \cos \beta \cos (\alpha -\beta )\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\]\[(2\cos \alpha \cos \beta -\cos (\alpha -\beta ))\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\]\[[\cos (\alpha +\beta )+\cos (\alpha -\beta )-\cos (\alpha -\beta )]\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\cos (\alpha +\beta )\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta -{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha (1-{{\cos }^{2}}\beta )-{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha {{\sin }^{2}}\beta -{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\sin }^{2}}\beta \] = 0.You need to login to perform this action.
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