If frac{2sin alpha}{1 + cos alpha + sin alpha | vedclass

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(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. Using half-angle identities $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alph

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$y$

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(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. Using half-angle identities $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ and $1 + \cos \alpha = 2\cos^2 \frac{\alpha}{2}$: $\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2})}{2\cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = y$ $\frac{4\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2\cos \frac{\alpha}{2}(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})} = y$ $\frac{2\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}} = y$. Now,consider the expression $E = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha}$. Using $1 - \cos \alpha = 2\sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ and $1 + \sin \alpha = (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2$: $E = \frac{2\sin^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}} = y$.

If $\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$,then $\frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} = $

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