Prove the following identities :

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

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{sec A + tan A}}$

Multiplying numerator and denominator by (sec A - tan A), we get :

$\Rightarrow \dfrac{1}{\text{sec A + tan A}} \times \dfrac{\text{sec A - tan A}}{\text{sec A - tan A}} \\[1em] \Rightarrow \dfrac{\text{sec A - tan A}}{\text{sec}^2 A - \text{tan}^2 A}$

By formula,

sec² A - tan² A = 1

$\Rightarrow \dfrac{\text{sec A - tan A}}{1} \\[1em] \Rightarrow \text{sec A - tan A}.$

Since, L.H.S. = R.H.S.

Hence proved that $\dfrac{1}{\text{sec A + tan A}}$ = sec A - tan A.