Question

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

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Solution

## Disclaimer: There is some error in the given question. The question should have been Question: If $a=\frac{2\mathrm{sin}x}{1+\mathrm{cos}x+\mathrm{sin}x}$, then prove that $\frac{1-\mathrm{cos}x+\mathrm{sin}x}{1+\mathrm{sin}\mathit{}x}$ is also equal to a. So, the solution is done accordingly. Solution: $a=\frac{2\mathrm{sin}x}{1+\mathrm{sin}x+\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\mathrm{Rationalising}\mathrm{the}\mathrm{denominator}:\phantom{\rule{0ex}{0ex}}\frac{2\mathrm{sin}x}{1+\mathrm{sin}x+\mathrm{cos}x}×\frac{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x}{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}x\left\{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x\right\}}{{\left(1+\mathrm{sin}x\right)}^{2}-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}x\left\{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x\right\}}{1+{\mathrm{sin}}^{2}x+2\mathrm{sin}x-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}x\left\{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x\right\}}{2{\mathrm{sin}}^{2}x+2\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}x\left\{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x\right\}}{2\mathrm{sin}x\left(1+\mathrm{sin}x\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x}{1+\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}\therefore a=\frac{\left(1+\mathrm{sin}x\right)-\mathrm{cos}x}{1+\mathrm{sin}x}$ Hence proved.

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