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I will go ahead and verify these trigonometric identities one by one. Let’s start with the first one and proceed sequentially.

### Identity 1:
\[
\text{Given: } \csc \theta - \sin \theta = \cot \theta \cos \theta
\]
To verify:

1. Recall that \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).
2. So the left-hand side (LHS) becomes:
   \[
   \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta}.
   \]
3. The right-hand side (RHS) is:
   \[
   \cot \theta \cos \theta = \frac{\cos \theta}{\sin \theta} \cdot \cos \theta = \frac{\cos^2 \theta}{\sin \theta}.
   \]
4. Therefore, LHS = RHS. Identity 1 is **verified**.

### Identity 2:
\[
\text{Given: } \sin x + \cos x \cdot \cot x = \csc x
\]
To verify:

1. Using \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\).
2. LHS becomes:
   \[
   \sin x + \cos x \cdot \frac{\cos x}{\sin x} = \sin x + \frac{\cos^2 x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x} = \frac{1}{\sin x}.
   \]
3. RHS is \(\csc x = \frac{1}{\sin x}\).
4. Therefore, LHS = RHS. Identity 2 is **verified**.

### Identity 3:
\[
\text{Given: } \frac{\sec^2 2u - 1}{\sec^2 2u} = \sin^2 2u
\]
To verify:

1. Using the identity \(\sec^2 \theta - 1 = \tan^2 \theta\).
2. The LHS becomes:
   \[
   \frac{\sec^2 2u - 1}{\sec^2 2u} = \frac{\tan^2 2u}{\sec^2 2u} = \sin^2 2u,
   \]
   because \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\) and \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\).
3. Therefore, LHS = RHS. Identity 3 is **verified**.

### Identity 4:
\[
\text{Given: } \tan t + 2 \cos t \cdot \csc t = \sec t \cdot \csc t + \cot t
\]
To verify:

1. Using \(\csc t = \frac{1}{\sin t}\), \(\sec t = \frac{1}{\cos t}\), and \(\cot t = \frac{\cos t}{\sin t}\).
2. LHS becomes:
   \[
   \tan t + 2 \cos t \cdot \frac{1}{\sin t} = \frac{\sin t}{\cos t} + \frac{2 \cos t}{\sin t} = \frac{\sin^2 t + 2 \cos^2 t}{\sin t \cdot \cos t}.
   \]
3. RHS becomes:
   \[
   \frac{1}{\cos t} \cdot \frac{1}{\sin t} + \frac{\cos t}{\sin t} = \frac{1}{\sin t \cdot \cos t} + \frac{\cos t}{\sin t} = \frac{1 + \cos^2 t}{\sin t \cdot \cos t}.
   \]
4. Both sides are equal, and thus Identity 4 is **verified**.

### Identity 5:
\[
\text{Given: } \frac{\csc^2 \theta}{1 + \tan^2 \theta} = \cot^2 \theta
\]
To verify:

1. Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\) and \(\csc^2 \theta = 1 + \cot^2 \theta\).
2. LHS becomes:
   \[
   \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{1 + \cot^2 \theta}{\frac{1}{\cos^2 \theta}} = \cot^2 \theta.
   \]
3. Therefore, LHS = RHS. Identity 5 is **verified**.

I'll continue with the other identities, but before that, do you want detailed verification steps for all of them or just final confirmations? Let me know how you'd like to proceed! 

Here are some follow-up questions to deepen your understanding:
1. Can you derive the identity \(\csc^2 \theta = 1 + \cot^2 \theta\) from first principles?
2. How would you explain the relationship between \(\sec \theta\) and \(\cos \theta\)?
3. Why is the identity \(1 + \tan^2 \theta = \sec^2 \theta\) so useful in verifying trigonometric identities?
4. Can you graphically represent how \(\tan \theta\) behaves compared to \(\sin \theta\) and \(\cos \theta\)?
5. What is the significance of rewriting trigonometric functions in terms of \(\sin\) and \(\cos\)?

**Tip:** When solving trigonometric identities, it's helpful to express all functions in terms of \(\sin\) and \(\cos\) for simplification.

Verify the following trigonometric identities.

This guide provides a detailed verification of several trigonometric identities, including the use of Pythagorean, reciprocal, and quotient identities. Suitable for high school students studying advanced mathematics.

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