A) \[0\]
B) \[1\]
C) \[{{\alpha }^{2}}-{{\beta }^{2}}\]
D) \[{{\alpha }^{2}}+{{\beta }^{2}}\]
Correct Answer: A
Solution :
Let\[\Delta =\left| \begin{matrix} 1 & \cos (\alpha -\beta ) & \cos \alpha \\ \cos (\alpha -\beta ) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{matrix} \right|\] Applying\[{{R}_{2}}\to {{R}_{2}}-\cos (\alpha -\beta ){{R}_{1}}\], \[{{R}_{3}}\to {{R}_{3}}-\cos \alpha {{R}_{1}}\] \[\left| \begin{matrix} 1 & \cos (\alpha -\beta ) \\ 0 & 1-{{\cos }^{2}}(\alpha -\beta ) \\ 0 & \cos \beta -\cos \alpha \cos (\alpha -\beta ) \\ \end{matrix} \right.\] \[\left. \begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \alpha \\ & \cos \beta \cos \alpha \cos (\alpha -\beta ) \\ & \,\,\,\,\,\,\,\,\,\,\,1-{{\cos }^{2}}\alpha \\ \end{align} \right|\] \[=[1-{{\cos }^{2}}(\alpha -\beta )][1-{{\cos }^{2}}\alpha ]\] \[-{{[\cos \beta -\cos \alpha (\alpha -\beta )]}^{2}}\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}(\alpha -\beta )+{{\cos }^{2}}\alpha {{\cos }^{2}}(\alpha -\beta )\] \[-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha {{\cos }^{2}}(\alpha -\beta )+2\cos \alpha \cos \beta \] \[\cos (\alpha -\beta )\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -{{\cos }^{2}}(\alpha -\beta )\] \[+2\cos \alpha \cos \beta \cos (\alpha -\beta )\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta )[\cos (\alpha -\beta )\] \[-2\cos \alpha \cos \beta ]\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta )\] \[[\cos (\alpha -\beta )-\cos (\alpha +\beta )-\cos (\alpha -\beta )]\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta )\] \[[-\cos (\alpha +\beta )]\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta +\cos (\alpha -\beta )\cos (\alpha +\beta )\] \[=1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta +{{\cos }^{2}}\alpha -{{\sin }^{2}}\beta \] \[=1-({{\cos }^{2}}\beta +{{\sin }^{2}}\beta )=1-1\] \[=0\]
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