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

(sin A + sec A)² + (cos A + cosec A)² = (1 + sec A cosec A)².

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

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

Since, L.H.S. = R.H.S. hence proved that, (sin A + sec A)² + (cos A + cosec A)² = (1 + sec A cosec A)².