Consider the following two statements.Stateme | vedclass

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(A) For statement $p$: We have $\sin 120^\circ = \frac{\sqrt{3}}{2}$,so $2 \sin 120^\circ = \sqrt{3}$.Substituting $	heta = 240^\circ$ in the $RHS$:

Vedclass · Accepted Answer

$F, T$

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(A) For statement $p$: We have $\sin 120^\circ = \frac{\sqrt{3}}{2}$,so $2 \sin 120^\circ = \sqrt{3}$.Substituting $	heta = 240^\circ$ in the $RHS$: $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}}$.Since $\sqrt{2 \pm \sqrt{3}} = \sqrt{\frac{4 \pm 2\sqrt{3}}{2}} = \frac{\sqrt{3} \pm 1}{\sqrt{2}}$,the expression becomes $\frac{\sqrt{3}-1}{2} - \frac{\sqrt{3}+1}{2} = -1 
eq \sqrt{3}$. Thus,statement $p$ is False.For statement $q$: In any quadrilateral $ABCD$,$A + B + C + D = 360^\circ = 2\pi$. Therefore,$\frac{A+C}{2} + \frac{B+D}{2} = \pi$. Let $\alpha = \frac{A+C}{2}$,then $\frac{B+D}{2} = \pi - \alpha$. The equation becomes $\cos \alpha + \cos(\pi - \alpha) = \cos \alpha - \cos \alpha = 0$. Thus,statement $q$ is True.Therefore,the truth values are $F, T$.

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