To prove the given trigonometric identity:

$\frac{\cot y (1 + \tan^2 y)}{\csc y} = \sec y$

Let's start by simplifying the left-hand side (LHS) of the equation.

We know the following trigonometric identities:

1. $\cot y = \frac{\cos y}{\sin y}$
2. $\tan y = \frac{\sin y}{\cos y}$
3. $\csc y = \frac{1}{\sin y}$
4. $\sec y = \frac{1}{\cos y}$
5. $1 + \tan^2 y = \sec^2 y$ (Pythagorean identity)

Using these identities, we can simplify the LHS:

$\frac{\cot y (1 + \tan^2 y)}{\csc y} = \frac{\frac{\cos y}{\sin y} (1 + \tan^2 y)}{\frac{1}{\sin y}}$

Simplify the denominator:

$= \frac{\frac{\cos y}{\sin y} (1 + \tan^2 y) \cdot \sin y}{1}$

The $\sin y$ in the numerator and denominator cancels out:

$= \cos y (1 + \tan^2 y)$

Using the Pythagorean identity $1 + \tan^2 y = \sec^2 y$:

$= \cos y \cdot \sec^2 y$

Since $\sec y = \frac{1}{\cos y}$, we can write:

$= \cos y \cdot \left(\frac{1}{\cos y}\right)^2$

Simplify the expression:

$= \cos y \cdot \frac{1}{\cos^2 y} = \frac{\cos y}{\cos^2 y} = \frac{1}{\cos y} = \sec y$

Thus, we have shown that:

$\frac{\cot y (1 + \tan^2 y)}{\csc y} = \sec y$

This completes the proof.

Would you like more details or have any questions?

Here are five more questions you might consider:

1. How do you prove the Pythagorean trigonometric identities?
2. What is the difference between $\cot y$ and $\tan y$?
3. How do you derive the identities for $\sec y$ and $\csc y$?
4. Can you solve an example problem involving these identities?
5. How can you apply these identities in calculus problems?

Tip: When dealing with trigonometric identities, always look for opportunities to use fundamental identities like Pythagorean identities to simplify the expressions.