Q188 / 270 · NDA 2017 (Phase 2) · Mathematics
$\sqrt{1+\sin \mathrm{A}}=-\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$ is true if
- A$\frac{3 \pi}{2} \lt \mathrm{A} \lt \frac{5 \pi}{2}$ only
- B$\frac{\pi}{2} \lt \mathrm{A} \lt \frac{3 \pi}{2}$ only
- Correct answerC$\frac{3 \pi}{2} \lt \mathrm{A} \lt \frac{7 \pi}{2}$
- D$0 \lt \mathrm{A} \lt \frac{3 \pi}{2}$
Solution
$\sqrt{1+\sin \mathrm{A}}=-\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)$
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,
$\frac{3 \pi}{4} \lt \frac{\mathrm{A}}{2} \lt \frac{7 \pi}{4}$
$\Rightarrow \frac{3 \pi}{2} \lt \mathrm{A} \lt \frac{7 \pi}{2}$.
Hence, the correct answer is option: 'C'.