The set of all possible values of theta in th | vedclass

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(D) Two points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of the line $ax+by+c=0$ if the expressions $(ax_1+by_1+c)$ and $(ax_2+by_2+c)$ have

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$\left(0, \frac{\pi}{2}\right)$

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(D) Two points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of the line $ax+by+c=0$ if the expressions $(ax_1+by_1+c)$ and $(ax_2+by_2+c)$ have the same sign,i.e.,$(ax_1+by_1+c)(ax_2+by_2+c) > 0$.Given the line $x+y-1=0$ and points $(1, 2)$ and $(\sin heta, \cos heta)$.First,evaluate the expression at $(1, 2)$:$1+2-1 = 2$,which is positive.Therefore,for the points to lie on the same side,the expression at $(\sin heta, \cos heta)$ must also be positive:$\sin heta + \cos heta - 1 > 0$$\Rightarrow \sin heta + \cos heta > 1$Divide by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin heta + \frac{1}{\sqrt{2}} \cos heta > \frac{1}{\sqrt{2}}$$\Rightarrow \sin \left( heta + \frac{\pi}{4} ight) > \sin \left(\frac{\pi}{4} ight)$Since $ heta \in (0, \pi)$,then $ heta + \frac{\pi}{4} \in \left(\frac{\pi}{4}, \frac{5\pi}{4} ight)$.In this interval,$\sin \left( heta + \frac{\pi}{4} ight) > \frac{1}{\sqrt{2}}$ holds when:$\frac{\pi}{4} Subtracting $\frac{\pi}{4}$ from all parts:$0 < heta < \frac{\pi}{2}$

The set of all possible values of $\theta$ in the interval $(0, \pi)$ for which the points $(1, 2)$ and $(\sin \theta, \cos \theta)$ lie on the same side of the line $x+y=1$ is

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