Prove that :

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

To prove:

Equation : $\dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}}$.

Solving L.H.S. of the above equation :

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

Solving R.H.S. of the above equation :

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

Since, L.H.S. = R.H.S = 1 + cos A.

Hence, proved that $\dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}}$.