Question

If $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2},$ then $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is equal to _________.

Answer 1

If $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}$ is given,
$\begin{array}{rcl}\sqrt{\frac{1-\sin x}{1+\sin x}}& =& \sqrt{\frac{{\sin }^2{\displaystyle \frac{x}{2}}+{\cos }^2{\displaystyle \frac{x}{2}}-2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{{\sin }^2{\displaystyle \frac{x}{2}}+{\cos }^2{\displaystyle \frac{x}{2}}+2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}}\\ & & \left(\mathrm{using}\mathrm{identics}:-{\sin }^2\theta +{\cos }^2\theta -1\sin 2\theta =2\sin \theta \cos \theta \right)\\ & =& \sqrt{\frac{{\left(\sin {\displaystyle \frac{x}{2}}-\cos {\displaystyle \frac{x}{2}}\right)}^2}{{\left(\sin {\displaystyle \frac{x}{2}}+\cos {\displaystyle \frac{x}{2}}\right)}^2}}\\ & =& \frac{\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}}\\ & =& \frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}+1}\left(\mathrm{dividing}\mathrm{by}\cos \frac{x}{2}\right)\\ & =& \frac{\tan {\displaystyle \frac{x}{2}}-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan {\displaystyle \frac{x}{2}}}\\ & =& \tan \left(\frac{x}{2}-\frac{\mathrm{\pi }}{4}\right)\end{array}$
$\mathrm{i}.\mathrm{e}\sqrt{\frac{1-\sin x}{1+\sin x}}=\tan \left(\frac{x}{2}-\frac{\mathrm{\pi }}{4}\right)$