Simplify the following expression:
$$\frac{{\cos A}}{{1 - \tan A}} + \frac{{\sin A}}{{1 - \cot A}} - \sin A$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{\cos A}}{{1 - \tan A}} + \frac{{\sin A}}{{1 - \cot A}} - \sin A \cr & = \frac{{\cos A}}{{1 - \frac{{\sin A}}{{\cos A}}}} + \frac{{\sin A}}{{1 - \frac{{\cos A}}{{\sin A}}}} - \sin A \cr & = \frac{{{{\cos }^2}A}}{{\cos A - \sin A}} + \frac{{{{\sin }^2}A}}{{\sin A - \cos A}} - \sin A \cr & = \frac{{{{\cos }^2}A}}{{\cos A - \sin A}} - \frac{{{{\sin }^2}A}}{{\cos A - \sin A}} - \sin A \cr & = \frac{{{{\cos }^2}A - {{\sin }^2}A}}{{\cos A - \sin A}} - \sin A \cr & = \frac{{{{\cos }^2}A - {{\sin }^2}A - \sin A\cos A + {{\sin }^2}A}}{{\cos A - \sin A}} \cr & = \frac{{{{\cos }^2}A - \sin A\cos A}}{{\cos A - \sin A}} \cr & = \frac{{\cos A\left( {\cos A - \sin A} \right)}}{{\cos A - \sin A}} \cr & = \boxed{\cos A} \cr & \cr & {\bf{Alternative: - }} \cr & {\text{You can put any }}A{\text{ value and satisfied equation,}} \cr & {\text{let }}A = {135^ \circ } \cr & \frac{{\cos {{135}^ \circ }}}{{1 - \tan {{135}^ \circ }}} + \frac{{\sin {{135}^ \circ }}}{{1 - \cot {{135}^ \circ }}} - \sin {135^ \circ } \cr & = \frac{{ - \frac{1}{{\sqrt 2 }}}}{2} + \frac{{ - \frac{1}{{\sqrt 2 }}}}{2} - \frac{1}{{\sqrt 2 }} \cr & = - \frac{1}{{\sqrt 2 }} \cr & {\text{Put in option (D) }}\cos {135^ \circ } = - \frac{1}{{\sqrt 2 }}{\text{ satisfied}} \cr & \therefore {\text{Option D is correct}}{\text{.}} \cr} $$