If $$x = \frac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }},$$     then the value of $$\frac{{1 - \cos \theta + \sin \theta }}{{1 + \sin \theta }}$$   is

Solution (By Examveda Team)

$$\eqalign{ & x = \frac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }} \cr & {\text{Let's put the value of }}\theta = {45^ \circ } \cr & {\text{So, }}x = \frac{{2 \times \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}} \cr & \left\{ {{\text{Because }}\sin {{45}^ \circ } = \frac{1}{{\sqrt 2 }}{\text{ and}}\cos {{45}^ \circ } = \frac{1}{{\sqrt 2 }}} \right\} \cr & x = \frac{{\sqrt 2 }}{{\frac{{\left( {\sqrt 2 + 2} \right)}}{{\sqrt 2 }}}} \cr & x = \frac{{\sqrt 2 }}{{\left( {1 + \sqrt 2 } \right)}} \cr & {\text{Now, check in the required value}} \cr & \frac{{1 - \cos \theta + \sin \theta }}{{1 + \sin \theta }} \cr & = \frac{{1 - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }}}} \cr & = \frac{{\sqrt 2 }}{{\left( {\sqrt 2 + 1} \right)}} \cr & {\text{This values is same as }}x. \cr & {\text{Hence answer is }}x. \cr} $$