If $$a=2sin$$⁡θ$$1+$$$$cos$$⁡θ$$+$$$$sin$$⁡θ, then $$1+$$$$sin$$⁡θ−$$cos$$⁡θ$$1+$$$$sin$$⁡θ is equal to

We have, $$a=2sin$$⁡θ$$1+$$$$cos$$⁡θ$$+$$$$sin$$⁡θ⇒ $$a=2sin$$⁡θ(1+$$$$sin$$⁡θ−$$cos$$⁡θ$$)(1+$$$$cos$$⁡θ$$+$$$$sin$$⁡θ$$)(1+$$$$sin$$⁡θ−$$cos$$⁡θ$$)⇒ $$a=2sin$$⁡θ$$(1+$$$$sin$$⁡θ−$$cos$$⁡θ$$)(1+$$$$sin$$⁡θ$$)2$$−$$cos2$$⁡θ⇒ $$a=2sin$$⁡θ$$(1+$$$$sin$$⁡θ−$$cos$$⁡θ$$)2sin$$⁡θ$$(1+$$$$sin$$⁡θ$$)=1+$$$$sin$$⁡θ−$$cos$$⁡θ$$1+$$$$sin$$⁡θ

Questions

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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\frac{1}{a}

a

1 - a

1 + a

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detailed solution

Correct option is B

We have,

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

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