If alpha lies in the second quadrant,then sqr | vedclass

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Question

(A) Let the expression be $E = \sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}} - \sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}$.Rationalizing the denominators

Vedclass · Accepted Answer

$2 	an \alpha$

Vedclass · Answer

(A) Let the expression be $E = \sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}} - \sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}$.Rationalizing the denominators:$E = \frac{1-\sin \alpha}{\sqrt{1-\sin^2 \alpha}} - \frac{1+\sin \alpha}{\sqrt{1-\sin^2 \alpha}} = \frac{1-\sin \alpha - (1+\sin \alpha)}{\sqrt{\cos^2 \alpha}}$.$E = \frac{-2 \sin \alpha}{|\cos \alpha|}$.Since $\alpha$ lies in the second quadrant,$\cos \alpha$ is negative,so $|\cos \alpha| = -\cos \alpha$.Thus,$E = \frac{-2 \sin \alpha}{-\cos \alpha} = 2 	an \alpha$.

If $\alpha$ lies in the second quadrant,then $\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}-\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}=$

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