Question

If $\alpha \in [\frac{\pi }{2},\pi ],$ then the value of $\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }$ is

• A. $2\cos \frac{\alpha }{2}$
• B. $2\sin \frac{\alpha }{2}$
• C. $2$
• D. $2\tan \frac{\alpha }{2}$

Answer 1

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

$\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }=\sqrt{{(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})}^2}-\sqrt{{(\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2})}^2}=|\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}|-|\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2}|\cdots \left(1\right)$

Now, $\alpha \in [\frac{\pi }{2},\pi ]$
$\Rightarrow \frac{\alpha }{2}\in [\frac{\pi }{4},\frac{\pi }{2}]\therefore \sin \frac{\alpha }{2}>\cos \frac{\alpha }{2}\Rightarrow \sin \frac{\alpha }{2}-\cos \frac{\alpha }{2}>0\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }=\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}=2\cos \frac{\alpha }{2}$