Prove the following identities :

sec² A . cosec² A = tan² A + cot² A + 2

Solving L.H.S. of the equation :

⇒ sec² A . cosec² A

⇒ $\dfrac{1}{\text{cos}^2 A. \text{ sin}^2 A}$.

Solving R.H.S. of the equation :

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

By formula,

sin² A + cos² A = 1.

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

Since, L.H.S. = R.H.S. = $\dfrac{1}{\text{cos}^2 A \text{ sin}^2 A}.$

Hence, proved that sec² A . cosec² A = tan² A + cot² A + 2.