For what values of theta are the points (1, 1 | vedclass

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(B) The points are collinear if the area of the triangle formed by them is $0$.$\frac{1}{2} \left| \begin{matrix} 1 & 1 & 1 \ 0 & \sec^2 	heta & 1 \

Vedclass · Accepted Answer

$	heta 
eq \frac{n\pi}{2}$

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(B) The points are collinear if the area of the triangle formed by them is $0$.$\frac{1}{2} \left| \begin{matrix} 1 & 1 & 1 \ 0 & \sec^2 	heta & 1 \ \csc^2 	heta & 0 & 1 \end{matrix} ight| = 0$Expanding the determinant:$1(\sec^2 	heta - 0) - 1(0 - \csc^2 	heta) + 1(0 - \sec^2 	heta \csc^2 	heta) = 0$$\sec^2 	heta + \csc^2 	heta - \sec^2 	heta \csc^2 	heta = 0$$\frac{1}{\cos^2 	heta} + \frac{1}{\sin^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$$\frac{\sin^2 	heta + \cos^2 	heta}{\sin^2 	heta \cos^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$$\frac{1}{\sin^2 	heta \cos^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$$0 = 0$This identity holds for all $	heta$ where $\sec^2 	heta$ and $\csc^2 	heta$ are defined.Since $\sec^2 	heta$ is undefined at $	heta = \frac{n\pi}{2}$ (where $\cos 	heta = 0$) and $\csc^2 	heta$ is undefined at $	heta = n\pi$ (where $\sin 	heta = 0$),the points are collinear for all $	heta 
eq \frac{n\pi}{2}$.

For what values of $\theta$ are the points $(1, 1), (0, \sec^{2}\theta), (\csc^{2}\theta, 0)$ collinear?

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