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If $y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}$, then value of $\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$ is
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$y$
$\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$
$=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}$
$=\frac{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{\left(1+\sin ^{2} \alpha+2 \sin \alpha\right)-\left(1-\sin ^{2} \alpha\right)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y$
$=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}$
$=\frac{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{\left(1+\sin ^{2} \alpha+2 \sin \alpha\right)-\left(1-\sin ^{2} \alpha\right)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y$
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