Question

If $x$ lies in $I{I}^{nd}$ quadrant, then $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$ is equal to

• A. $\sin \frac{x}{2}$
• B. $\tan \frac{x}{2}$
• C. $\sec \frac{x}{2}$
• D. $\csc \frac{x}{2}$

Answer 1

The correct option is B $\tan \frac{x}{2}$

We have,

$\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin \frac{x}{2}\cos \frac{x}{2}}+\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin \frac{x}{2}\cos \frac{x}{2}}}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin \frac{x}{2}\cos \frac{x}{2}}-\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin \frac{x}{2}\cos \frac{x}{2}}}$

$=\frac{\sqrt{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}+\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}}{\sqrt{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}-\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}}$

$=\frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}$

$=\frac{2\sin \frac{x}{2}}{2\cos \frac{x}{2}}$

$=\tan \frac{x}{2}$

Hence, this is the answer.