Trigonometry

Prove that: $\;$ $\dfrac{\cos A}{1 + \sin A} + \tan A = \sec A$


$\begin{aligned} LHS & = \dfrac{\cos A}{1 + \sin A} + \tan A \\\\ & = \dfrac{\cos A}{1 + \sin A} + \dfrac{\sin A}{\cos A} \\\\ & = \dfrac{\cos^2 A + \sin^2 A + \sin A}{\cos A \left(1 + \sin A\right)} \\\\ & = \dfrac{1 + \sin A}{\cos A \left(1 + \sin A\right)} \;\;\; \left[\text{Note: } \sin^2 A + \cos^2 A = 1\right] \\\\ & = \dfrac{1}{\cos A} \\\\ & = \sec A = RHS \end{aligned}$

Hence proved.