Consider three points P(cos alpha, sin beta), | vedclass

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(D) To check if the points $P(\cos \alpha, \sin \beta)$, $Q(\sin \alpha, \cos \beta)$, and $R(0,0)$ are collinear, we calculate the area of the triang

Vedclass · Accepted Answer

$P, Q, R$ are non-collinear

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(D) To check if the points $P(\cos \alpha, \sin \beta)$, $Q(\sin \alpha, \cos \beta)$, and $R(0,0)$ are collinear, we calculate the area of the triangle formed by them using the determinant method:$\Delta = \frac{1}{2} \left| \begin{array}{ccc} \cos \alpha & \sin \beta & 1 \ \sin \alpha & \cos \beta & 1 \ 0 & 0 & 1 \end{array} ight|$Expanding along the third row:$\Delta = \frac{1}{2} [1 \cdot (\cos \alpha \cos \beta - \sin \alpha \sin \beta)]$$\Delta = \frac{1}{2} \cos(\alpha + \beta)$Given $0 Since $\cos(\alpha + \beta) 
eq 0$ for $0 Therefore, the points $P, Q, R$ are non-collinear.

Consider three points $P(\cos \alpha, \sin \beta)$, $Q(\sin \alpha, \cos \beta)$, and $R(0,0)$, where $0 < \alpha, \beta < \frac{\pi}{4}$. Then:

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