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

$\dfrac{\text{cot A - 1}}{2 - \text{sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}$.

The L.H.S of the equation can be written as,

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

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