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

$\dfrac{1}{\text{sec A + tan A}} - \dfrac{1}{\text{cos A}} = \dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sec A - tan A}}$

The equation can be written as,

$\dfrac{1}{\text{sec A + tan A}} + \dfrac{1}{\text{sec A - tan A}} = \dfrac{2}{\text{cos A}}$

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

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

Since, L.H.S. = R.H.S. hence proved that,

$\dfrac{1}{\text{sec A + tan A}} - \dfrac{1}{\text{cos A}} = \dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sec A - tan A}}$.