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Question: \(\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} =\) (when x li...

1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x=\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = (when x lies in IInd quadrant)

A

sin⁡x2\sin\frac{x}{2}

B

tan⁡x2\tan\frac{x}{2}

C

sec⁡x2\sec\frac{x}{2}

D

cosecx2\text{cosec}\frac{x}{2}

Answer

tan⁡x2\tan\frac{x}{2}

Explanation

Solution

1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x=cos⁡x2+sin⁡x2+sin⁡x2−cos⁡x2cos⁡x2+sin⁡x2−sin⁡x2+cos⁡x2\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \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⁡x2= \tan\frac{x}{2}.