if a=2sin⁡θ1+cos⁡θ+sin⁡θ, then 1+sin⁡θ−cos⁡θ1+sin⁡θis

given a=2sin⁡θ1+cos⁡θ+sin⁡θ=2sinθ(1+sinθ−cosθ)(1+sinθ+cosθ)(1+sinθ−cosθ)=2sinθ(1+sinθ−cosθ)(1+sinθ)2−cos2θ=2sinθ(1+sinθ−cosθ)2sin2θ+2sinθ=1+sin⁡θ−cos⁡θ1+sin⁡θ=a

if a=2sin⁡θ1+cos⁡θ+sin⁡θ, then 1+sin⁡θ−cos⁡θ1+sin⁡θis

given a=2sin⁡θ1+cos⁡θ+sin⁡θ=2sinθ(1+sinθ−cosθ)(1+sinθ+cosθ)(1+sinθ−cosθ)=2sinθ(1+sinθ−cosθ)(1+sinθ)2−cos2θ=2sinθ(1+sinθ−cosθ)2sin2θ+2sinθ=1+sin⁡θ−cos⁡θ1+sin⁡θ=a

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if $a = \frac{2 \sin θ}{1 + \cos θ + \sin θ}, then \frac{1 + \sin θ - \cos θ}{1 + \sin θ}$ is

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1-a

1+a

1/a

answer is C.

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Detailed Solution

given $a = \frac{2 \sin θ}{1 + \cos θ + \sin θ}$
$\begin{array}{l} = \frac{2 \sin θ (1 + \sin θ - \cos θ)}{(1 + \sin θ + \cos θ) (1 + \sin θ - \cos θ)} \\ = \frac{2 \sin θ (1 + \sin θ - \cos θ)}{(1 + \sin θ)^{2} - \cos^{2} θ} \\ = \frac{2 \sin θ (1 + \sin θ - \cos θ)}{2 \sin^{2} θ + 2 \sin θ} \\ = \frac{1 + \sin θ - \cos θ}{1 + \sin θ} = a \end{array}$