A) Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
B) Statement 1 is true; Statement 2 is true; Statement 2 is a correct explanation for Statement 1.
C) Statement 1 is true; Statement 2 is false.
D) Statement 1 is false; Statement 2 is true.
Correct Answer: C
Solution :
\[{{\Delta }_{1}}=\left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & \cos \alpha \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 0 & \sin \alpha -\cos \alpha & \cos \alpha -\sin \alpha \\ 0 & \cos \alpha +\sin \alpha & \sin \alpha -\cos \alpha \\ 1 & -\sin \alpha & \cos \alpha \\ \end{matrix} \right|\] \[={{(\sin \alpha -\cos \alpha )}^{2}}-({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )\] \[={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha -2\sin \alpha ,\cos \alpha -{{\cos }^{2}}\alpha \] \[+{{\sin }^{2}}\alpha \] \[=2{{\sin }^{2}}\alpha -2\sin \alpha .\cos \alpha \] \[=2\sin \alpha (\sin \alpha -\cos \alpha )\] Now, \[\alpha -\cos \alpha =0\]for only\[\alpha =\frac{\pi }{4}\]in\[\left( 0,\frac{\pi }{2} \right)\] \[\therefore \]\[{{\Delta }_{1}}=2(\sin \alpha )\times 0=0,\] since value of sin a is finite for \[\alpha \in \left( 0,\frac{\pi }{2} \right)\] Hence non-trivivial solution for only one value of \[\alpha \] in\[\left( 0,\frac{\pi }{2} \right)\] \[\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} 0 & \sin \alpha & \cos \alpha \\ 0 & \cos \alpha & \sin \alpha \\ 2\cos \alpha & -\sin \alpha & -\cos \alpha \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[2\cos \alpha ({{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )=0\] \[\therefore \]\[\cos \alpha =0\]or\[{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha =0\] But \[\cos \alpha =0\] not possible for any value of\[\alpha \in \left( 0,\frac{\pi }{2} \right)\] \[\therefore \]\[{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha =0\Rightarrow \sin \alpha =-\cos \alpha ,\]which is also not possible for any value of\[\alpha \in \left( 0,\frac{\pi }{2} \right)\] Hence, there is no solution.You need to login to perform this action.
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