frac{{sqrt {1 + sin x} + sqrt {1 - sin x} }}{ | vedclass

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Question

(b) $\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}$

$= \frac{{\cos \frac{x}{2} + \sin \frac{x}{2} +

Vedclass · Accepted Answer

$	an \frac{x}{2}$

Vedclass · Answer

(b) $\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}$

$= \frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}}}$

$ = 	an \frac{x}{2}$.

$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} , \,\,($ जब $x \, \in $ द्वितीय चतुर्थांष $) =$

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