If theta lies in the second quadrant,then the | vedclass

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(B) Let $E = \sqrt{\frac{1 - \sin 	heta}{1 + \sin 	heta}} + \sqrt{\frac{1 + \sin 	heta}{1 - \sin 	heta}}$.Simplifying the expression:$E = \frac{\s

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$-2 \sec 	heta$

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(B) Let $E = \sqrt{\frac{1 - \sin 	heta}{1 + \sin 	heta}} + \sqrt{\frac{1 + \sin 	heta}{1 - \sin 	heta}}$.Simplifying the expression:$E = \frac{\sqrt{1 - \sin 	heta} \cdot \sqrt{1 - \sin 	heta} + \sqrt{1 + \sin 	heta} \cdot \sqrt{1 + \sin 	heta}}{\sqrt{(1 + \sin 	heta)(1 - \sin 	heta)}}$$E = \frac{(1 - \sin 	heta) + (1 + \sin 	heta)}{\sqrt{1 - \sin^2 	heta}}$$E = \frac{2}{\sqrt{\cos^2 	heta}} = \frac{2}{|\cos 	heta|}$.Since $	heta$ lies in the second quadrant,$\cos 	heta$ is negative.Therefore,$|\cos 	heta| = -\cos 	heta$.Thus,$E = \frac{2}{-\cos 	heta} = -2 \sec 	heta$.

If $\theta$ lies in the second quadrant,then the value of $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$ is:

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