Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{1 + tan A}}{\text{sin A}} + \dfrac{\text{1 + cot A}}{\text{cos A}} = \text{2(sec A + cosec A)}$

The L.H.S. of the equation can be written as,

$\Rightarrow \dfrac{1 + \dfrac{\text{sin A}}{\text{cos A}}}{\text{sin A}} + \dfrac{1 + \dfrac{\text{cos A}}{\text{sin A}}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos A}\Big(1 + \dfrac{\text{sin A}}{\text{cos A}}\Big) + \text{sin A}\Big(1 + \dfrac{\text{cos A}}{\text{sin A}}\Big)}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{cos A + sin A + sin A + cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{sin A cos A}} \\[1em] \Rightarrow 2\Big(\dfrac{\text{cos A}}{\text{sin A cos A}} + \dfrac{\text{sin A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow 2\Big(\dfrac{1}{\text{sin A}} + \dfrac{1}{\text{cos A}}\Big) \\[1em] \Rightarrow 2\text{(cosec A + sec A)}.$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{1 + tan A}}{\text{sin A}} + \dfrac{\text{1 + cot A}}{\text{cos A}}$ = 2(sec A + cosec A).