If $P(\sin \alpha, \cos \alpha)$ lies inside the triangle formed by the vertices $(0,0),\left(\sqrt{\frac{3}{2}}, 0\right)$ and $\left(0, \sqrt{\frac{3}{2}}\right)$, then $\alpha$ lies in the interval

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Aicharya · Accepted Answer

$P$ and $A$ will lie on the same side of $x=0$. $\therefore \quad \sin \alpha>0$ ...(i) $P$ and $\dot{B}$ will lie on the same side of $y=0$. $\therefore \quad \cos \alpha>0$ ...(ii) $P$ and $O$ will lie on the same side of $x+y=\sqrt{\frac{3}{2}}$. $\begin{aligned} & \therefore\left(0+0-\frac{\sqrt{3}}{2}\right)\left(\sin \alpha+\cos \alpha-\frac{\sqrt{3}}{2}\right)>0 \\ & \Rightarrow\left(0+0-\sqrt{\frac{3}{2}}\right)\left(\sin \alpha+\cos \alpha-\sqrt{\frac{3}{2}}\right)>0 \\ & \Rightarrow \sin \alpha+\cos \alpha < \sqrt{\frac{3}{2}} \Rightarrow \sin \left(\alpha+\frac{\pi}{4}\right) < \frac{\sqrt{3}}{2}\end{aligned}$ $\alpha+\frac{\pi}{4} \in\left(0, \frac{\pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \pi\right)$ ...(iii) From Eqs. (i), (ii) and (iii), $\alpha \in\left(0, \frac{\pi}{12}\right) \cup\left(\frac{5 \pi}{12}, \frac{\pi}{2}\right)$

If $P(\sin \alpha, \cos \alpha)$ lies inside the triangle formed by the vertices $(0,0)…

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