Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}.$

Solving L.H.S.,

$\Rightarrow \dfrac{\text{cos A}}{1 - \dfrac{\text{sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos A}}{\dfrac{\text{cos A - sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos A - sin A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{(cos A - sin A)(cos A + sin A)}}{\text{(cos A - sin A)}} \\[1em] = \text{cos A + sin A}.$

Since, L.H.S. = R.H.S. hence proved that $\dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}$.