Question

If $x=\frac{2\sin x}{1+\cos x+\sin x}$, then prove that $\frac{1-\cos x+\sin x}{1+\sin \mathit{}x}$ is also equal to a.

Answer 1

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 $\frac{1-\cos x+\sin x}{1+\sin \mathit{}x}$ is also equal to a.
So, the solution is done accordingly.

Solution:
$$\begin{gathered} a=\frac{2\sin x}{1+\sin x+\cos x} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{2\sin x}{1+\sin x+\cos x}\times \frac{\left(1+\sin x\right)-\cos x}{\left(1+\sin x\right)-\cos x} \\ =\frac{2\sin x\left\{\left(1+\sin x\right)-\cos x\right\}}{{\left(1+\sin x\right)}^2-{\cos }^{2}x} \\ =\frac{2\sin x\left\{\left(1+\sin x\right)-\cos x\right\}}{1+{\sin }^{2}x+2\sin x-{\cos }^{2}x} \\ =\frac{2\sin x\left\{\left(1+\sin x\right)-\cos x\right\}}{2{\sin }^{2}x+2\sin x} \\ =\frac{2\sin x\left\{\left(1+\sin x\right)-\cos x\right\}}{2\sin x\left(1+\sin x\right)} \\ =\frac{\left(1+\sin x\right)-\cos x}{1+\sin x} \\ \therefore a=\frac{\left(1+\sin x\right)-\cos x}{1+\sin x} \end{gathered}$$

Hence proved.