Question

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

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

Answer 1

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

Given

$\alpha \in [\frac{\pi }{2},\pi ]$

$⟹\frac{\alpha }{2}\in [\frac{\pi }{4},\frac{\pi }{2}]$

$⟹-\frac{\alpha }{2}\in [-\frac{\pi }{2},-\frac{\pi }{4}]$

$⟹\frac{\pi }{4}-\frac{\alpha }{2}\in [-\frac{\pi }{4},0]$

$⟹\cos (\frac{\pi }{4}-\frac{\alpha }{2})>0$ and $\sin (\frac{\pi }{4}-\frac{\alpha }{2})<0$

$\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }$

$⟹\sqrt{1+\cos (\frac{\pi }{2}-\alpha )}-\sqrt{1-\cos (\frac{\pi }{2}-\alpha )}$

$(\sin \theta =\cos (\frac{\pi }{2}-\cos \theta ))$

$⟹\sqrt{2{\cos }^2(\frac{\pi }{4}-\frac{\alpha }{2})}-\sqrt{2{\sin }^2(\frac{\pi }{4}-\frac{\alpha }{2})}$

$(1+\cos \theta =2{\cos }^2\theta ,1-\cos \theta =2{\sin }^2\theta )$

$⟹\sqrt{2}\left|\cos (\frac{\pi }{4}-\frac{\alpha }{2})\right|-\sqrt{2}\left|\sin (\frac{\pi }{4}-\frac{\alpha }{2})\right|$

$⟹\sqrt{2}(\cos (\frac{\pi }{4}-\frac{\alpha }{2})+\sin (\frac{\pi }{4}-\frac{\alpha }{2}))$

$(|x|=\{\begin{array}{c}xx\ge 0\\ -xx>0\end{array})$

$⟹\sqrt{2}(\cos \frac{\pi }{4}\cos \frac{\alpha }{2}+\sin \frac{\pi }{2}\sin \frac{\alpha }{2}+\sin \frac{\pi }{4}\cos \frac{\alpha }{2}-\cos \frac{\pi }{4}\sin \frac{\alpha }{2})$

$(\cos (A-B)=\cos A\cos B+\sin A\sin B,\sin (A-B)=\sin A\cos B-\cos A\sin B)$

$⟹\sqrt{2}(\frac{\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}}{\sqrt{2}}+\frac{\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}}{\sqrt{2}})$

$⟹\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}$

$⟹2\cos \frac{\alpha }{2}$