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

sec⁴ A(1 - sin⁴ A) - 2 tan² A = 1.

Solving L.H.S.,

$\Rightarrow \dfrac{1}{\text{cos}^4 A}\text{(1 + sin }^2 A)\text{(1 - sin}^2 A) - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{(1 + sin}^2 A)\text{cos}^2 A}{{\text{cos}^4 A}} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A}{\text{cos}^2 A} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A - 2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 - \text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos}^2 A} \\[1em] = 1.$

Since, L.H.S. = R.H.S. hence proved that sec⁴ A(1 - sin⁴ A) - 2tan² A = 1.