If 2sin⁡θ1+cos⁡θ+sin⁡θ=y, then 1−cos⁡θ+sin⁡θ1+sin⁡θ=

Given x=2sin⁡θ1+cos⁡θ+sin⁡θ=2sin⁡θ1+sin⁡θ+cos⁡θ×1+sin⁡θ−cos⁡θ1+sin⁡θ−cos⁡θ=2sin⁡θ(1+sin⁡θ−cos⁡θ)2sin⁡θ(1+sin⁡θ) ⇒1−cos⁡θ+sin⁡θ1+sin⁡θ=y

If 2sin⁡θ1+cos⁡θ+sin⁡θ=y, then 1−cos⁡θ+sin⁡θ1+sin⁡θ=

Given x=2sin⁡θ1+cos⁡θ+sin⁡θ=2sin⁡θ1+sin⁡θ+cos⁡θ×1+sin⁡θ−cos⁡θ1+sin⁡θ−cos⁡θ=2sin⁡θ(1+sin⁡θ−cos⁡θ)2sin⁡θ(1+sin⁡θ) ⇒1−cos⁡θ+sin⁡θ1+sin⁡θ=y

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$If \frac{2 \sin θ}{1 + \cos θ + \sin θ} = y, then \frac{1 - \cos θ + \sin θ}{1 + \sin θ} =$

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$y$

$\frac{1}{y}$

$1 - y$

$1 + y$

answer is A.

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Detailed Solution

$Given x = \frac{2 \sin θ}{1 + \cos θ + \sin θ} = \frac{2 \sin θ}{1 + \sin θ + \cos θ} \times \frac{1 + \sin θ - \cos θ}{1 + \sin θ - \cos θ} = \frac{2 \sin θ (1 + \sin θ - \cos θ)}{2 \sin θ (1 + \sin θ)} \Rightarrow \frac{1 - \cos θ + \sin θ}{1 + \sin θ} = y$