Collinearity of Three Points Defined by Trigonometric Functions

To determine the collinearity of the three points $P, Q,$ and $R$, we analyze their coordinates using vector algebra and determinant properties.

Let the given points be represented as:
$P = (x_1, y_1) = (-\sin(\beta-\alpha), -\cos \beta)$
$Q = (x_2, y_2) = (\cos(\beta-\alpha), \sin \beta)$
$R = (x_3, y_3) = (\cos(\beta-\alpha+\theta), \sin(\beta-\theta))$

Step 1: Express $R$ in terms of $P$ and $Q$

Using the trigonometric angle sum and difference formulas for $x_3$ and $y_3$:

For $x_3$:
$x_3 = \cos(\beta-\alpha+\theta) = \cos(\beta-\alpha)\cos\theta - \sin(\beta-\alpha)\sin\theta$
Substituting $x_2 = \cos(\beta-\alpha)$ and $x_1 = -\sin(\beta-\alpha)$:
$x_3 = x_2 \cos\theta + x_1 \sin\theta$

For $y_3$:
$y_3 = \sin(\beta-\theta) = \sin\beta\cos\theta - \cos\beta\sin\theta$
Substituting $y_2 = \sin\beta$ and $y_1 = -\cos\beta$:
$y_3 = y_2 \cos\theta + y_1 \sin\theta$

Step 2: Formulate the Collinearity Determinant

Three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear if and only if the determinant $\Delta$ formed by their coordinates is zero:

$$\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

Substitute the relations $x_3 = x_1 \sin\theta + x_2 \cos\theta$ and $y_3 = y_1 \sin\theta + y_2 \cos\theta$ into the third row:

$$\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_1 \sin\theta + x_2 \cos\theta & y_1 \sin\theta + y_2 \cos\theta & 1 \end{vmatrix}$$

Apply the row operation $R_3 \to R_3 - (\sin\theta) R_1 - (\cos\theta) R_2$:

$$\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ 0 & 0 & 1 - \sin\theta - \cos\theta \end{vmatrix}$$

Expanding along the third row yields:
$\Delta = (1 - \sin\theta - \cos\theta)(x_1 y_2 - x_2 y_1)$

Step 3: Evaluate $x_1 y_2 - x_2 y_1$

Using the definitions of $x_1, y_1, x_2, y_2$:
$x_1 y_2 - x_2 y_1 = (-\sin(\beta-\alpha))(\sin\beta) - (\cos(\beta-\alpha))(-\cos\beta)$
$= \cos(\beta-\alpha)\cos\beta - \sin(\beta-\alpha)\sin\beta$

Applying the cosine addition identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$x_1 y_2 - x_2 y_1 = \cos((\beta-\alpha) + \beta) = \cos(2\beta - \alpha)$

Thus, the determinant simplifies to:
$\Delta = (1 - \sin\theta - \cos\theta)\cos(2\beta - \alpha)$

Step 4: Check if $\Delta = 0$

Given the range $0 < \alpha, \beta, \theta < \frac{\pi}{4}$:

1. For the factor $(1 - \sin\theta - \cos\theta)$:
   $\sin\theta + \cos\theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)$
   Since $0 < \theta < \frac{\pi}{4}$, we have $\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{\pi}{2}$, which gives:
   $1 < \sin\theta + \cos\theta < \sqrt{2}$
   Therefore, $(1 - \sin\theta - \cos\theta) \neq 0$.

2. For the factor $\cos(2\beta - \alpha)$:
   Since $0 < \alpha < \frac{\pi}{4}$ and $0 < \beta < \frac{\pi}{4}$:
   $0 < 2\beta < \frac{\pi}{2} \implies -\frac{\pi}{4} < 2\beta - \alpha < \frac{\pi}{2}$
   In the interval $\left(-\frac{\pi}{4}, \frac{\pi}{2}\right)$, the cosine function is strictly positive, so $\cos(2\beta - \alpha) > 0$.

Since neither factor is zero, $\Delta \neq 0$.

Conclusion

Because the determinant is non-zero, the area of $\triangle PQR$ is non-zero, which means that the points $P, Q, R$ are non-collinear.

Correct Answer: (D) $P, Q, R$ are non-collinear