Q. If$\ x\$lies in IInd quadrant, then$\ \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) $cosec\ \frac{x}{2}$
Explanation
(b)
We have$,\ \ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}$
$= \frac{\sqrt{\left( \sin{\frac{x}{2} + \cos\frac{x}{2}} \right)^{2}} + \sqrt{\left( \sin{\frac{x}{2} - \cos\frac{x}{2}} \right)^{2}}}{\sqrt{\left( \sin{\frac{x}{2} + \cos\frac{x}{2}} \right)^{2}} - \sqrt{\left( \sin{\frac{x}{2} - \cos\frac{x}{2}} \right)^{2}}}$
$= \frac{\cos{\frac{x}{2} + \sin{\frac{x}{2} + \sin{\frac{x}{2} - \cos\frac{x}{2}}}}}{\cos{\frac{x}{2} + \sin{\frac{x}{2} - \sin{\frac{x}{2} + \cos\frac{x}{2}}}}} = \tan\frac{x}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.