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We are asked to prove the following trigonometric identity:

\[
\frac{1 - \cos \theta}{1 + \cos \theta} = 2 \csc^2 \theta - 2 \csc \theta \cot \theta - 1
\]

### Step-by-Step Solution:

1. **Left-hand Side (LHS):**  
   Start with the left-hand side (LHS):

   \[
   \frac{1 - \cos \theta}{1 + \cos \theta}
   \]

   To simplify this, we can use the identity for sine and cosine in terms of tangent:
   
   \[
   \frac{1 - \cos \theta}{1 + \cos \theta} = \left(\frac{\sin^2 \theta}{(1 + \cos \theta)} \right) 
   \]
   
   This uses the identity \(\sin^2 \theta = 1 - \cos^2 \theta\).

2. **Convert to Terms of \(\sin \theta\) and \(\cos \theta\):**
   Next, let’s rewrite the terms on the right-hand side in terms of \(\sin \theta\) and \(\cos \theta\).

   The right-hand side (RHS) is:
   
   \[
   2 \csc^2 \theta - 2 \csc \theta \cot \theta - 1
   \]
   
   Using the definitions:
   \[
   \csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}
   \]
   
   Substituting these into the RHS:
   \[
   2 \csc^2 \theta = \frac{2}{\sin^2 \theta}
   \]
   and
   \[
   2 \csc \theta \cot \theta = \frac{2 \cos \theta}{\sin^2 \theta}
   \]
   
   So the RHS becomes:
   \[
   \frac{2}{\sin^2 \theta} - \frac{2 \cos \theta}{\sin^2 \theta} - 1
   \]

3. **Combine Terms:**
   Combine the terms on the right-hand side into a single fraction:
   \[
   \frac{2 - 2 \cos \theta}{\sin^2 \theta} - 1
   \]

   Rewrite \(1\) as \(\frac{\sin^2 \theta}{\sin^2 \theta}\):
   \[
   \frac{2 - 2 \cos \theta - \sin^2 \theta}{\sin^2 \theta}
   \]

4. **Use Pythagorean Identity:**
   Now, use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), so \(\sin^2 \theta = 1 - \cos^2 \theta\).

   Substituting this into the expression, we get:
   \[
   \frac{2 - 2 \cos \theta - (1 - \cos^2 \theta)}{\sin^2 \theta}
   \]

   Simplifying the numerator:
   \[
   2 - 2 \cos \theta - 1 + \cos^2 \theta = 1 - 2 \cos \theta + \cos^2 \theta
   \]

   Therefore, the RHS becomes:
   \[
   \frac{1 - 2 \cos \theta + \cos^2 \theta}{\sin^2 \theta}
   \]

5. **Factor and Conclude:**
   Notice that the expression \(1 - 2 \cos \theta + \cos^2 \theta\) is a perfect square:
   \[
   (1 - \cos \theta)^2
   \]

   So the expression becomes:
   \[
   \frac{(1 - \cos \theta)^2}{\sin^2 \theta}
   \]

   Finally, recognize that:
   \[
   \frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta
   \]

Thus, both sides are equal, proving the identity.

---

Do you want more details or further clarification? Here are five related questions you might find interesting:

1. How can you use trigonometric identities to simplify other expressions like this one?
2. What is the importance of the Pythagorean identity in trigonometry?
3. How do the reciprocal identities like \(\csc \theta\) and \(\cot \theta\) relate to simplifications?
4. Are there any geometric interpretations for identities like this one?
5. What are the common strategies for proving trigonometric identities?

**Tip:** When proving trigonometric identities, converting everything to sine and cosine can often simplify the process.

Prove the trigonometric identity: (1 - cos θ) / (1 + cos θ) = 2 csc² θ - 2 csc θ cot θ - 1

Explore how to prove the trigonometric identity (1 - cos θ) / (1 + cos θ) = 2 csc² θ - 2 csc θ cot θ - 1. This problem uses key concepts from trigonometry such as reciprocal functions and the Pythagorean identity. Ideal for Grades 10-12 students.

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