$$\frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{1 + \sin A + \cos A}}$$    is equal to:

Solution(By Examveda Team)

In these type of questions we can go through option-
From option B
$$\eqalign{ & \frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right) + \cos A}} = 1 - \cos A + \sin A \cr & 2\sin A\left( {1 + \sin A} \right) = {\left( {1 + \sin A} \right)^2} - {\cos ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 1 + \sin 2A + 2\sin A - 1 + {\sin ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 2\sin A\left( {1 + \sin A} \right) \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put }}A = {90^ \circ } \cr & \frac{{\left( {2\sin {{90}^ \circ }} \right)\left( {1 + \sin {{90}^ \circ }} \right)}}{{1 + \sin {{90}^ \circ } + \cos {{90}^ \circ }}} = 2 \cr & {\text{Now from option B}} \cr & 1 + \sin A - \cos A \cr & = 1 + \sin {90^ \circ } - \cos {90^ \circ } \cr & = 1 + 1 - 0 \cr & = 2 \cr} $$