If P $\equiv$ $(-sin(\beta - \alpha), -cos\beta)$, Q $\equiv$ $(cos(\beta - \alpha), sin\beta)$ and R $\equiv$ $(cos(\beta - \alpha + \theta), sin(\beta - \theta))$, where $$0 \lt \alpha, \beta, \theta \lt \frac{\pi}{4}$$ then show that P, Q and R are non- collinear.
I tried my best but could not solve this so I looked at the solution. The solution reads:
P $\equiv$ $(-sin(\beta - \alpha), -cos\beta)$ $\equiv$ $(x_1, y_1)$
Q $\equiv$ $(cos(\beta - \alpha), sin\beta)$ $\equiv (x_2, y_2)$
R $\equiv$ $(cos(\beta - \alpha + \theta), sin(\beta - \theta))$ $ \equiv (x_2cos\theta + x_1sin\theta, y_2cos\theta +y_1sin\theta )$
If $$ T \equiv (\frac{x_2cos\theta + x_1sin\theta}{cos\theta + sin\theta}, \frac{y_2cos\theta + y_1sin\theta}{cos\theta + sin\theta})$$
then P, Q, T are collinear. Hence P, Q and R are noncollinear
If $Z$ is on the line determined by $X,Y$ then $Z=aX+bY$ and $a+b=1$
In your question, in order for $R$ to be colinear with $P,Q$ we'll need $$cos(\theta)+sin(\theta)=1$$
$$cos(\theta)+sin(\theta)=\sqrt{2}sin(\theta+\pi/4)$$ because $sin$ function monotonically increase on $[0,\pi/2]$, $\sqrt{2}sin(0+\pi/4)=1$, $\sqrt{2}sin(\pi/4+\pi/4)=\sqrt{2}$ there is no $\theta$ in the given range satisfies the condition.