If the point (2 cos theta, 2 sin theta) for t | vedclass

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Question

(C) The point $(2 \cos 	heta, 2 \sin 	heta)$ lies on the circle $x^2 + y^2 = 4$. We are given the lines $x+y=2$ and $x-y=2$. The region containing t

Vedclass · Accepted Answer

$\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

Vedclass · Answer

(C) The point $(2 \cos 	heta, 2 \sin 	heta)$ lies on the circle $x^2 + y^2 = 4$. We are given the lines $x+y=2$ and $x-y=2$. The region containing the origin is defined by the inequalities $x+y Substituting $x = 2 \cos 	heta$ and $y = 2 \sin 	heta$: $1$) $2 \cos 	heta + 2 \sin 	heta $2$) $2 \cos 	heta - 2 \sin 	heta From the provided figure, the shaded region corresponds to the part of the circle where the $x$-coordinate is less than $0$ (i.e.,$\cos 	heta Thus, the correct option is $C$.

If the point $(2 \cos \theta, 2 \sin \theta)$ for $\theta \in (0, 2 \pi)$ lies in the region between the lines $x+y=2$ and $x-y=2$ containing the origin, then $\theta$ lies in

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