Question

If $\alpha \in \left[\frac{\pi }{2}\pi \right]$ then the value of $\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }$ is equal to:

• A. $2\cos \frac{\alpha }{2}$
• B. $2\sin \frac{\alpha }{2}$
• C. $2$
• D. none of these

Answer 1

The correct option is A $2\cos \frac{\alpha }{2}$

$\alpha \in [\frac{\pi }{2},\pi ]$
$\Rightarrow \frac{\alpha }{2}\in [\frac{\pi }{4},\frac{\pi }{2}]$
$\therefore \sin \alpha >0,\cos \alpha <0$
$\sqrt{1+\sin \alpha }=\sqrt{(\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}{)}^2}$
$=\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}$ (As $\cos \frac{\alpha }{2}\ge 0,\sin \frac{\alpha }{2}>0$)
$\sqrt{1-\sin \alpha }=\sqrt{(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}{)}^2}$
$=\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2}$ (As $\sin \frac{\alpha }{2}\ge \cos \frac{\alpha }{2}$)
$\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }=\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}-\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}$
$=2\cos \frac{\alpha }{2}$