1+ A =- ( A 2 + A 2 ) is true if

$\sqrt{1+\sin \mathrm{A}}=-$\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$We know,$1+\sin \mathrm{A}=$\left(\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}\right)^2$ $\therefore \sqrt{1+\sin \mathrm{A}}=\left|\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}\right|$ We know, |x|=\left\{$\begin{array}{c}x \text{ if } x \geq 0 \\ -x \text{ if } x \lt 0\end{array}$\right. $\left|\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right|=$\begin{cases}\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}, & \text{ if } 2 \mathrm{n} \pi-\frac{\pi}{4} \leq \frac{\mathrm{A}}{2} \leq 2 \mathrm{n} \pi+\frac{3 \pi}{4} \\ -\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right), & \text{ if otherwise }\end{cases} \therefore \sqrt{1+\sin \mathrm{A}}=-$\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$ when, $\frac{3 \pi}{4} \lt \frac{\mathrm{A}}{2} \lt \frac{7 \pi}{4}$ $\Rightarrow \frac{3 \pi}{2} \lt \mathrm{A} \lt \frac{7 \pi}{2}$.