If $x=\frac{2 \sin x}{1+\cos x+\sin x}$, then prove that $\frac{1-\cos x+\sin x}{1+\sin x}$ is also equal to $a$.
Disclaimer: There is some error in the given question.
The question should have been
Question: If $a=\frac{2 \sin x}{1+\cos x+\sin x}$, then prove that $\frac{1-\cos x+\sin x}{1+\sin x}$ is also equal to $a .$
So, the solution is done accordingly.
Solution:
$a=\frac{2 \sin x}{1+\sin x+\cos x}$
Rationalising the denominator :
$\frac{2 \sin x}{1+\sin x+\cos x} \times \frac{(1+\sin x)-\cos x}{(1+\sin x)-\cos x}$
$=\frac{2 \sin x\{(1+\sin x)-\cos x\}}{(1+\sin x)^{2}-\cos ^{2} x}$
$=\frac{2 \sin x\{(1+\sin x)-\cos x\}}{1+\sin ^{2} x+2 \sin x-\cos ^{2} x}$
$=\frac{2 \sin x\{(1+\sin x)-\cos x\}}{2 \sin ^{2} x+2 \sin x}$
$=\frac{2 \sin x\{(1+\sin x)-\cos x\}}{2 \sin x(1+\sin x)}$
$=\frac{(1+\sin x)-\cos x}{1+\sin x}$
$\therefore a=\frac{(1+\sin x)-\cos x}{1+\sin x}$
Hence proved.