A) \[\theta =\frac{n\pi }{2}\]
B) \[\theta \ne \frac{n\pi }{2}\]
C) \[\theta =n\pi \]
D) None of these
Correct Answer: B
Solution :
The given points are collinear, if Area \[=\frac{1}{2}\,\left| \,\begin{matrix} 1 & 1 & 1 \\ 0 & {{\sec }^{2}}\theta & 1 \\ \text{cose}{{\text{c}}^{2}}\theta & 0 & 1 \\ \end{matrix}\, \right|=0\] \[\Rightarrow \,\,1({{\sec }^{2}}\theta )+1(\text{cose}{{\text{c}}^{2}}\theta )-1(\text{cose}{{\text{c}}^{2}}\theta .{{\sec }^{2}}\theta )=0\] \[\Rightarrow \,\,\frac{1}{{{\cos }^{2}}\theta }+\frac{1}{{{\sin }^{2}}\theta }-\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0\] \[\Rightarrow \,\,\frac{1}{{{\cos }^{2}}\theta {{\sin }^{2}}\theta }-\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0\,\,\,\Rightarrow \,\,0=0\] Therefore, the points are collinear for all values of q, except only \[\theta =\frac{n\pi }{2}\] because at \[\theta =\frac{n\pi }{2},\,\,{{\sec }^{2}}\theta =\infty .\]
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