If frac{2 sin theta}{1+cos theta+sin theta}=y | vedclass

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Question

(A) Given $y = \frac{2 \sin 	heta}{1+\cos 	heta+\sin 	heta}$.Using half-angle formulas $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}

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$y$

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

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

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