Prove the following identity:

$\Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A$

Solving L.H.S of the equation:

$\Rightarrow \dfrac{\sin A \times \dfrac{\sin A}{\cos A}}{1 - \cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A(1 - \cos A)} \\[1em] \text{ By formula, } \sin^2 A = 1 - \cos^2 A \\[1em] \Rightarrow \dfrac{1 - \cos^2 A}{\cos A(1 - \cos A)} \\[1em] \Rightarrow \dfrac{(1 + \cos A)(1 - \cos A)}{\cos A(1 - \cos A)} \\[1em] \Rightarrow \dfrac{(1 + \cos A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} + \dfrac{\cos A}{\cos A} \\[1em] \Rightarrow \sec A + 1.$

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

Hence, proved that $\Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A$.