Assertion (A): If sqrt{4 sin^4 theta + sin^2 | vedclass

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(C) Given equation: $\sqrt{4 \sin^4 	heta + \sin^2 2	heta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{	heta}{2}ight) = 2$. Using $\sin 2	heta = 2 \si

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$(A)$ is true but $(R)$ is false

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(C) Given equation: $\sqrt{4 \sin^4 	heta + \sin^2 2	heta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{	heta}{2}ight) = 2$. Using $\sin 2	heta = 2 \sin 	heta \cos 	heta$,the first term is $\sqrt{4 \sin^4 	heta + 4 \sin^2 	heta \cos^2 	heta} = \sqrt{4 \sin^2 	heta (\sin^2 	heta + \cos^2 	heta)} = \sqrt{4 \sin^2 	heta} = 2 |\sin 	heta|$. Using $2 \cos^2 x = 1 + \cos 2x$,the second term is $2(1 + \cos(\frac{\pi}{2} - 	heta)) = 2(1 + \sin 	heta)$. Substituting these back: $2 |\sin 	heta| + 2 + 2 \sin 	heta = 2$,which simplifies to $|\sin 	heta| + \sin 	heta = 0$. This holds true if $\sin 	heta \leq 0$,which means $	heta$ lies in the $3^{	ext{rd}}$ or $4^{	ext{th}}$ quadrant. Thus,$(A)$ is true. For the reason $(R)$,$\sqrt{\sin^2 	heta} = |\sin 	heta|$,not $\sin 	heta$. Thus,$(R)$ is false.

Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant.
Reason $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$

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Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant. Reason $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$