Find the value of 
$$\frac{2 + \tan^2 2A + \cot^2 2A}{\sec 2A \cdot \cosec 2A}$$  = ? 

Correct option is A

Given:

Simplify the following expression:

​$$\frac{2 + \tan^2 2A + \cot^2 2A}{\sec 2A \cdot \cosec 2A}$$ 

Formula Used:

Using the identities $$\tan^2 \theta + 1 = \sec^2 \theta$$​ and $$\cot^2 \theta + 1 = \cosec^2 \theta$$ ,

​$$\tan^2 2A = \sec^2 2A - 1\\\ \\\cot^2 2A = \cosec^2 2A - 1$$ 

Solution:

Substituting these into the original expression gives:

​$$\frac{2 + (\sec^2 2A - 1) + (\cosec^2 2A - 1)}{\sec 2A \cdot \cosec 2A}\\ \ \\ = \frac{2 + \sec^2 2A + \cosec^2 2A - 2}{\sec 2A \cdot \cosec 2A}\\\ \\= \frac{\sec^2 2A + \cosec^2 2A}{\sec 2A \cdot \cosec 2A}$$​​

= $$\frac{\sec^2 2A}{\sec 2A \cdot \cosec 2A} + \frac{\cosec^2 2A}{\sec 2A \cdot \cosec 2A}$$ 

= $$\frac{\sec 2A}{\cosec 2A} + \frac{\cosec 2A}{ \sec 2A}$$​​

= tan2A + cot 2A