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

Question

If $y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}$, then value of $\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$ is, $\frac{y}{3}$, $\mathrm{y}$, $2 y$, $\frac{3}{2} y$

MARKS App · Accepted Answer

$\begin{aligned} & \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \ & =\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha} \ & =\frac{(1+\sin \alpha)^2-\cos ^2 \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)} \ & =\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)} \ & =\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y \end{aligned}$

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