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$\sqrt{1+\sin \mathrm{A}}=-\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$ is true if
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The correct answer is:
$\frac{3 \pi}{2} < \mathrm{A} < \frac{7 \pi}{2}$
$\sqrt{1+\sin A}=-\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)$
We know, $1+\sin A=\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)^{2}$
$\therefore \sqrt{1+\sin A}=\left|\cos \frac{A}{2}+\sin \frac{A}{2}\right|$
We know, $|x|=\left\{\begin{array}{l}x \text { if } x \geq 0 \\ -x \text { if } x < 0\end{array}\right.$
$\sin \frac{A}{2}+\cos \frac{A}{2} \mid=\left\{\begin{array}{l}\sin \frac{A}{2}+\cos \frac{A}{2}, \quad \text { if } 2 n \pi-\frac{\pi}{4} \leq \frac{A}{2} \leq 2 n \pi+\frac{3 \pi}{4} \\ -\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right),\end{array}\right.$
$\quad \therefore \sqrt{1+\sin A}=-\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)$ when
$\frac{3 \pi}{4} < \frac{A}{2} < \frac{7 \pi}{4}$
$\Rightarrow \frac{3 \pi}{2} < A < \frac{7 \pi}{2}$
We know, $1+\sin A=\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)^{2}$
$\therefore \sqrt{1+\sin A}=\left|\cos \frac{A}{2}+\sin \frac{A}{2}\right|$
We know, $|x|=\left\{\begin{array}{l}x \text { if } x \geq 0 \\ -x \text { if } x < 0\end{array}\right.$
$\sin \frac{A}{2}+\cos \frac{A}{2} \mid=\left\{\begin{array}{l}\sin \frac{A}{2}+\cos \frac{A}{2}, \quad \text { if } 2 n \pi-\frac{\pi}{4} \leq \frac{A}{2} \leq 2 n \pi+\frac{3 \pi}{4} \\ -\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right),\end{array}\right.$
$\quad \therefore \sqrt{1+\sin A}=-\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)$ when
$\frac{3 \pi}{4} < \frac{A}{2} < \frac{7 \pi}{4}$
$\Rightarrow \frac{3 \pi}{2} < A < \frac{7 \pi}{2}$
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