A) \[\theta =n\phi ,n=0,1,2,....\]
B) \[\theta +\phi =n\pi ,n=0,1,2,....\]
C) \[\theta =\phi +(2n+1)\frac{\pi }{2},n=0,1,2,....\]
D) \[\theta =\phi +n\frac{\pi }{2},n=0,1,2,....\]
E) \[\theta =\phi +3n\frac{\pi }{2},n=0,1,2,....\]
Correct Answer: C
Solution :
\[\therefore \]\[AB=\left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \cos \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\] \[\left[ \begin{matrix} {{\cos }^{2}}\phi & \sin \phi \cos \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\sin \theta \cos \theta \cos \phi \sin \phi \\ {{\cos }^{2}}\phi \cos \theta \sin \theta +{{\sin }^{2}}\theta \sin \phi \cos \phi \\ \end{matrix} \right.\] \[\left. \begin{matrix} {{\cos }^{2}}\theta \sin \phi \cos \phi +{{\sin }^{2}}\phi \sin \theta \cos \theta \\ \cos \theta \sin \theta \sin \phi \cos \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos \theta \cos \phi \cos (\theta -\phi ) \\ \sin \theta \cos \phi \cos (\theta -\phi ) \\ \end{matrix} \right.\] \[\left. \begin{matrix} \sin \phi \cos \theta \cos (\theta -\phi ) \\ \sin \theta \sin \phi \cos (\theta -\phi ) \\ \end{matrix} \right]\] \[\because \] \[AB=0\] \[\Rightarrow \] \[\cos (\theta -\phi )=0\] \[\Rightarrow \] \[\cos (\theta -\phi )=\cos (2n+1)\frac{\pi }{2}\] \[\Rightarrow \] \[\theta =(2n+1)\frac{\pi }{2}+\phi \] where, \[n=0,1,\text{ }2,...\]
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