Let's break down and establish the trigonometric identity given in the problem:

$(1 + \csc \theta)(1 - \csc \theta) = -\cot^2 \theta$

Step 1: Use the Difference of Squares

The left-hand side is in the form of a product of two binomials, which is the difference of squares formula: $(a + b)(a - b) = a^2 - b^2$ Here, $a = 1$ and $b = \csc \theta$, so applying the formula: $(1 + \csc \theta)(1 - \csc \theta) = 1^2 - (\csc \theta)^2 = 1 - \csc^2 \theta$

Step 2: Simplify using a Trigonometric Identity

We know from trigonometric identities that: $1 - \csc^2 \theta = -\cot^2 \theta$

This is a standard identity derived from the Pythagorean identity for cotangent and cosecant: $\csc^2 \theta - \cot^2 \theta = 1 \quad \Rightarrow \quad 1 - \csc^2 \theta = -\cot^2 \theta$

Conclusion

Thus, we have established that: $(1 + \csc \theta)(1 - \csc \theta) = -\cot^2 \theta$ which proves the identity.

Do you want more details or further explanation on this?

Here are 5 related questions to explore further:

1. What is the Pythagorean identity involving $\csc \theta$ and $\cot \theta$?
2. How is the difference of squares formula applied in other trigonometric identities?
3. Can you derive $\csc^2 \theta - \cot^2 \theta = 1$ from basic definitions?
4. What are other common trigonometric identities involving $\csc \theta$ and $\cot \theta$?
5. How would this identity change for other values of $\theta$ in different quadrants?

Tip: Always look for opportunities to apply algebraic identities like the difference of squares to simplify complex trigonometric expressions.