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

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

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

$\Rightarrow \dfrac{{\dfrac{1}{\text{cos A}} - 1}}{\dfrac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 - \text{cos A}}{\text{cos A}}} {\dfrac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)} \times \text{cos A}}{\text{(1 + cos A)} \times \text{cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}.$

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