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Question
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
Solution
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.
\[a = \frac{2\sin x}{1 + \sin x + \cos x}\]
Rationalising the denominator:
\[\frac{2\sin x}{1 + \sin x + \cos x} \times \frac{\left( 1 + \sin x \right) - \cos x}{\left( 1 + \sin x \right) - \cos x}\]
\[ = \frac{2\sin x\left\{ \left( 1 + \sin x \right) - \cos x \right\}}{\left( 1 + \sin x \right)^2 - \cos^2 x}\]
\[ = \frac{2\sin x\left\{ \left( 1 + \sin x \right) - \cos x \right\}}{1 + \sin^2 x + 2\sin x - \cos^2 x}\]
\[ = \frac{2\sin x\left\{ \left( 1 + \sin x \right) - \cos x \right\}}{2 \sin^2 x + 2\sin x}\]
\[ = \frac{2\sin x\left\{ \left( 1 + \sin x \right) - \cos x \right\}}{2\sin x\left( 1 + \sin x \right)}\]
\[ = \frac{\left( 1 + \sin x \right) - \cos x}{1 + \sin x}\]
\[ \therefore a = \frac{\left( 1 + \sin x \right) - \cos x}{1 + \sin x}\]
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