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

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(B) Given expression: $E = \frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}$We know that $1 + \sin x = \c

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$\cot \frac{x}{2}$

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(B) Given expression: $E = \frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}$We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$And $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$Since $x$ lies in the $II^{nd}$ quadrant,$\frac{\pi}{2} In this interval,$\cos \frac{x}{2} > 0$ and $\sin \frac{x}{2} > 0$,and $\cos \frac{x}{2} > \sin \frac{x}{2}$.Thus,$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.Substituting these into the expression:$E = \frac{(\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} - \sin \frac{x}{2})}$$E = \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}$.

$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)

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