MCQ
The points $(1,{\rm{ }}1)$, $(0,{\sec ^2}\theta ),\,({\rm{cose}}{{\rm{c}}^2}\theta ,{\rm{ }}0)$ are collinear for
- A$\theta = \frac{{n\pi }}{2}$
- ✓$\theta \ne \frac{{n\pi }}{2}$
- C$\theta = n\pi $
- DNone of these
Area $ = \frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\0&{{{\sec }^2}\theta }&1\\{{\rm{cose}}{{\rm{c}}^2}\theta }&0&1\end{array}\,} \right| = 0$
$ \Rightarrow \,\,1({\sec ^2}\theta ) + 1({\rm{cose}}{{\rm{c}}^2}\theta ) - 1({\rm{cose}}{{\rm{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 $\theta$, except only $\theta = \frac{{n\pi }}{2}$ because at $\theta = \frac{{n\pi }}{2},\,\,{\sec ^2}\theta = \infty .$
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