Question
If $\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=\text{y},$ then prove that $\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}$ is also equal to y.
$\Big[\text{Hint: Express }\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}.\frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}\Big]$

Answer

We have, $\frac{2\sin\alpha}{1\cos\alpha+\sin\alpha}=\text{y}$
Now, $\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{(1-\cos\alpha+\sin\alpha)}{(1+\sin\alpha)}.\frac{(1+\cos\alpha+\sin\alpha)}{(1+\cos\alpha+\sin\alpha)}$
$=\frac{(1+\sin\alpha)^2-\cos^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{1+\sin^2\alpha+2\sin\alpha-1+\sin^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{2\sin\alpha(1+\sin\alpha)}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{2\sin\alpha}{1+\sin\alpha+\cos\alpha}=\text{y}$

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