(sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A

Solving L.H.S. of the above equation :

$\Rightarrow \text{sin A sec A + sin A cosec A + cos A sec A + cos A cosec A} \\[1em] \Rightarrow \text{sin A} \times \dfrac{1}{\text{cos A}} + \text{sin A} \times \dfrac{1}{\text{sin A}} + \text{cos A} \times \dfrac{1}{\text{cos A}} + \text{cos A} \times \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + 1 + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow 2 + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \text{cosec A sec A}.$

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

Hence, proved that (sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A.