If $x=\frac{2 sin θ}{(1+cos θ+sin θ)}$, then the value of $\frac{1−cos θ+sin θ}{(1+sin θ)}$ is:

$x$

We are given that :-

x =  \(\frac{ 2 sinθ}{ 1 + cosθ + sinθ }\)

Now multiply and divide RHS by ( 1 - cosθ + sinθ )

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1 + cosθ + sinθ) × ( 1 - cosθ + sinθ )}\)

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)²  - cos²θ}\)

{ we know , sin²θ + cos²θ = 1 }

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)²  - ( 1 - sin²θ) }\)

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)²  - ( 1 - sinθ).( 1+ sinθ) }\)

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)(1 + sinθ - 1+ sinθ) }\)

= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)(2 sinθ) }\)

= \(\frac{  ( 1 - cosθ + sinθ ) }{( 1+ sinθ) }\)

So, 

\(\frac{  ( 1 - cosθ + sinθ ) }{( 1+ sinθ) }\) = x