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  • Statement 1 : The system of linear equations <img alt="x+(\sin \alpha )y+(\cos \alpha )z=0" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cinline%20x&plu

Statement 1 : The system of linear equations

x+(\sin \alpha )y+(\cos \alpha )z=0

x+(\cos \alpha )y+(\sin \alpha )z=0

x-(\sin \alpha )y-(\cos \alpha )z=0 

Has non-trival soluction for only one value of \alpha lying in the interval \left ( 0,\frac{\pi }{2} \right ).

Statement 2 : 

The equastion in \alpha 

\begin{vmatrix} \cos \alpha &\sin \alpha &\cos \alpha \\ \sin \alpha &\cos \alpha &\sin \alpha \\ \cos \alpha &-\sin \alpha & -\cos \alpha \end{vmatrix} = 0

Has only one solution lying in the interval \left ( 0,\frac{\pi }{2} \right ).

Option: 1

Statement 1 is true ; ststement 2 is true ; statement 2 is not correct explanation for statement 1


Option: 2

Statement 1 is true ; ststement 2 is true ; statement 2 is  correct explanation for statement 1


Option: 3

Statement 1 is true ; ststement 2 is false ; 


Option: 4

Statement 1 is false ; ststement 2 is true; 


Answers (1)

best_answer

\\\Delta'=\left|\begin{array}{ccc} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & \cos \alpha \end{array}\right|\\ R_1\rightarrow R_1-R_2,\;R_2\rightarrow R_2-R_3 \\\Delta' =\left|\begin{array}{ccc} 0 & \sin \alpha-\cos \alpha & \cos \alpha-\sin \alpha \\ 0 & \cos \alpha+\sin \alpha & \sin \alpha-\cos \alpha \\ 1 & -\sin \alpha & \cos \alpha \end{array}\right|

\\\mathrm{\;\;\;}=(\sin \alpha-\cos \alpha)^{2}-\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)\\\mathrm{\;\;\;}=\sin ^{2} \alpha+\cos ^{2} \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^{2} \alpha+\sin^2\alpha\\\mathrm{\;\;\;} =2 \sin ^{2} \alpha-2 \sin \alpha \cdot \cos \alpha \\\mathrm{\;\;\;} =2 \sin \alpha(\sin \alpha-\cos \alpha)\\\text { Now, } \sin \alpha-\cos \alpha=0 \text { for only }

\\\alpha=\frac{\pi}{4} \text { in }\left(0, \frac{\pi}{2}\right) \\ \therefore \quad \Delta_{1}=2(\sin \alpha) \times 0=0

since the value of \sin \alpha is finite for \alpha \in\left(0, \frac{\pi}{2}\right). Hence non-trivial solution for only one value of \alpha \in \left(0, \frac{\pi}{2}\right)

Now consider,

\begin{array}{l} \left|\begin{array}{lll} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{ccc} 0 & \sin \alpha & \cos \alpha \\ 0 & \cos \alpha & \sin \alpha \\ 2 \cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0 \end{array}

\\\Rightarrow 2 \cos \alpha\left(\sin ^{2} \alpha-\cos ^{2} \alpha\right)=0 \\ \therefore \cos \alpha=0 \quad \text { or } \quad \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 posible for any value of \alpha \in\left(0, \frac{\pi}{2}\right).

 

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mansi

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