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

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

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

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

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