Prove the following identities :

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

Solving L.H.S. of the equation :

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

By formula,

sin² A = 1 - cos² A

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

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

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