If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)]

According to the question,

y =2sinα /(1+cosα+sinα)

Multiplying numerator and denominator by (1 – cos α + sin α),

We get,

$\Rightarrow y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha} \times \frac{1-\cos \alpha+\sin \alpha}{1-\cos \alpha+\sin \alpha}$

$=\frac{2 \sin \alpha}{(1+\sin \alpha)+\cos \alpha} \times \frac{(1+\sin \alpha)-\cos \alpha}{(1+\sin \alpha)-\cos \alpha}$

Using $(a+b)(a-b)=a^{2}-b^{2}$, we get:

$=\frac{2 \sin \alpha\{(1+\sin \alpha)-\cos \alpha\}}{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}$

$=\frac{2 \sin \alpha(1+\sin \alpha)-2 \sin \alpha \cos \alpha}{1+\sin ^{2} \alpha+2 \sin \alpha-\cos ^{2} \alpha}$

Since, $1-\cos ^{2} \alpha=\sin ^{2} \alpha$

$\therefore y=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{\sin ^{2} \alpha+2 \sin \alpha+\sin ^{2} \alpha}$

$=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{2 \sin \alpha(1+\sin \alpha)}$

$\Rightarrow y=\frac{(1+\sin \alpha-\cos \alpha)}{(1+\sin \alpha)}$

$\Rightarrow y=\frac{(1-\cos \alpha+\sin \alpha)}{(1+\sin \alpha)}$

Hence Proved