Evaluate the following expression in terms of trigonometric ratios.
$$\frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{1 + \sin A}}$$

Evaluate the following expression in terms of trigonometric ratios.
$$\frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{1 + \sin A}}$$ Correct Answer $$\frac{{{{\sec }^2}A\left( {1 - \sin A} \right)}}{{1 + \sec A}}$$

$$\eqalign{ & \frac{{{{\cot }^2}A\left( {\sec A - 1} \right)}}{{\left( {1 + \sin A} \right)}} \cr & = \frac{{{{\cos }^2}A\left( {1 - \cos A} \right)}}{{{{\sin }^2}A\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 - {{\sin }^2}A} \right)}}{{\left( {1 - {{\cos }^2}A} \right)}} \times \frac{{\left( {1 - \cos A} \right)}}{{\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right).\left( {1 - \cos A} \right)}}{{\left( {1 + \cos A} \right)\left( {1 - \cos A} \right).\left( {1 + \sin A} \right).\cos A}} \cr & = \frac{{\frac{{\sec A\left( {1 - \sin A} \right)}}{{\left( {\sec A + 1} \right)}}}}{{\sec A}} \cr & = \frac{{{{\sec }^2}A\left( {1 - \sin A} \right)}}{{\left( {1 + \sec A} \right)}} \cr} $$