Prove the following identities :

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

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos A}}{\text{1 - sin A}}$

Multiplying numerator and denominator by (1 + sin A)

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

By formula,

cos² A = 1 - sin² A

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

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

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