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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 equal to

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a

\frac{1}{a}

b

a

c

1 - a

d

1 + a

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detailed solution

Correct option is B

We have,

$a = \frac{2 \sin θ}{1 + \cos θ + \sin θ}$

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

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