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 deno

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{\sqrt{(1 - \sin \alpha)^2}}{\sqrt{1 - \sin^2 \alpha}} - \frac{\sqrt{(1 + \sin \alpha)^2}}{\sqrt{1 - \sin^2 \alpha}}$$E = \frac{|1 - \sin \alpha| - |1 + \sin \alpha|}{\sqrt{\cos^2 \alpha}}$$E = \frac{(1 - \sin \alpha) - (1 + \sin \alpha)}{|\cos \alpha|}$ (Since $\sin \alpha \in (0, 1)$ in the second quadrant,$1 - \sin \alpha > 0$ and $1 + \sin \alpha > 0$)$E = \frac{-2 \sin \alpha}{|\cos \alpha|}$.Since $\alpha$ is in the second quadrant,$\cos \alpha $E = \frac{-2 \sin \alpha}{-\cos \alpha} = 2 	an \alpha$.

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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