Let $${\text{a}} = \frac{{2\sin x}}{{1 + \sin x + \cos x}}$$    and $${\text{b}} = \frac{{\text{c}}}{{1 + \sin x}}.$$   Then a = b, if c = ?

$$\eqalign{ & b = \frac{c}{{1 + \sin x}} \cr & {\text{Go through option from option B}} \cr & b = \frac{{1 + \sin x - \cos x}}{{1 + \sin x}} \cr & = \frac{{{{\left( {1 + \sin x} \right)}^2} - {{\cos }^2}x}}{{1 + \sin x\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{1 + {{\sin }^2}x + 2{{\sin }^2}x - 1 + {{\sin }^2}x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2{{\sin }^2}x + 2\sin x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2\sin x\left( {\sin x + 1} \right)}}{{1 + \sin x + \cos x}} \cr & = \frac{{2\sin x}}{{1 + \sin x + \cos x}} \cr & = a \cr} $$