Prove the following identities :

$\dfrac{\text{1 + cos A}}{\text{1 - cos A}} = \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2}$

Solving R.H.S. of the equation :

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

By formula,

sin² A = 1 - cos² A

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

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

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