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प्रश्न
If `(2sinalpha)/(1 + cosalpha + sinalpha)` = y, then prove that `(1 - cosalpha + sinalpha)/(1 + sinalpha)` is also equal to y.
`["Hint": "Express" (1 - cosalpha + sinalpha)/(1 + sinalpha) = (1 - cosalpha + sinalpha)/(1 + sinalpha) * (1 + cosalpha + sinalpha)/(1 + cosalpha + sinalpha)]`
उत्तर
Given that: y = `(2sinalpha)/(1 + cosalpha + sinalpha)`
= `(2sinalpha)/(1 + cosalpha + sinalpha) xx (1 + sinalpha - cosalpha)/(1 + sinalpha - cosalpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/((1 + sinalpha)^2 - cos^2alpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/(1 + sin^2alpha + 2sinalpha - cos^2alpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/((1 + cosalpha)^2 + sin^2alpha + 2sinalpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/(sin^2alpha + sin^2alpha + 2sinalpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/(2sin^2alpha + sinalpha)`
= `(2sinalpha(1 - cosalpha + sinalpha))/(2sinalpha(1 + sinalpha))`
= `(1 - cosalpha + sinalpha)/(1 + sinalpha)`
∴ `(1 - cosalpha + sinalpha)/(1 + sinalpha)` = y.
Hence proved.
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