Question

2sin theta /1+cos theta+sin theta=y then 1-cos theta +sin theta /1+sin yheta is also y

Answer 1

$Given, \\ \\ \frac{1-cos\theta +sin\theta}{1+sin\theta} \\ \\ = \frac{1+sin\theta +cos\theta}{1 +sin\theta} \\ \\ = \frac{1+sin\theta -cos\theta}{1 +sin\theta} \ * \frac{1+sin\theta +cos\theta}{1+sin\theta + cos\theta}} \\ \\ = \frac{(1 +sin\theta)^{2} -cos^{2}\theta} {(1+sin\theta)^{2} + cos\theta(1+sin\theta)} \\ \\ = \frac{ 1 + sin^{2}\theta + 2sin\theta - cos^{2}\theta}{1 + sin^{2}\theta + 2sin\theta + cos\theta + sin\theta . cos\theta} \\ \\ = \frac{ sin^{2}\theta + cos^{2}\theta + 2sin\theta + sin^{2}\theta - cos^{2}\theta}{ 1 + sin^{2}\theta + 2sin\theta + cos\theta + sin\theta . cos\theta} \\ \\ = \frac{2sin^{2}\theta + 2sin\theta}{ ( (1 + sin\theta)^{2} + cos\theta(1+ sin\theta)} \\ \\ = \frac{2sin\theta(sin\theta + 1)}{(1 + sin\theta)(1 + sin\theta + cos\theta)} \\ \\ = \frac{2sin\theta}{(1 + sin\theta + cos\theta)} \\ \\ = y \\ \\ \textbf{Hence Proved}$

Answer 2

x=1+cosθ+sinθ2sinθ,

Multiplying x above and below by 1−cosθ+sinθ, we get

x=(1+sinθ)2−cos2θ2sinθ(1−cosθ+sinθ)=(1+sinθ)2−(1−sin2θ)2sinθ(1−cosθ+sinθ)

Putting 1−sin2θ=(1+sinθ)(1−sinθ), we get 

⇒x=(1+sinθ)(1+sinθ−1+sinθ)2sinθ(1−cosθ+sinθ)

⇒x=2sinθ2sinθ⋅1+sinθ1−cosθ+sinθ

⇒x=1+sinθ1−cosθ+sinθ