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Let's solve the given trigonometric equalities one by one.

### (a) \(\frac{\sin u}{\csc u} + \frac{\cos u}{\sec u} = 1\)

To simplify:

1. Recall that \(\csc u = \frac{1}{\sin u}\) and \(\sec u = \frac{1}{\cos u}\).
2. Therefore, \(\frac{\sin u}{\csc u} = \sin u \times \sin u = \sin^2 u\), and \(\frac{\cos u}{\sec u} = \cos u \times \cos u = \cos^2 u\).
3. So, the left-hand side becomes:
   \[
   \sin^2 u + \cos^2 u = 1 \quad (\text{since this is a Pythagorean identity}).
   \]
Thus, the equation is verified as true.

### (b) \((1 - \cos^2 x)(1 + \cot^2 x) = 1\)

1. \(1 - \cos^2 x = \sin^2 x\) (using the Pythagorean identity).
2. Also, recall that \(\cot^2 x = \frac{\cos^2 x}{\sin^2 x}\).
3. Substituting, the left-hand side becomes:
   \[
   \sin^2 x \times \left(1 + \frac{\cos^2 x}{\sin^2 x}\right) = \sin^2 x \times \frac{\sin^2 x + \cos^2 x}{\sin^2 x}.
   \]
4. Since \(\sin^2 x + \cos^2 x = 1\), we have:
   \[
   \sin^2 x \times \frac{1}{\sin^2 x} = 1.
   \]
Thus, the equation is true.

### (c) \(\sin t(\csc t - \sin t) = \cos^2 t\)

1. Recall that \(\csc t = \frac{1}{\sin t}\), so \(\csc t - \sin t = \frac{1}{\sin t} - \sin t = \frac{1 - \sin^2 t}{\sin t}\).
2. Using the identity \(1 - \sin^2 t = \cos^2 t\), we get:
   \[
   \frac{\cos^2 t}{\sin t}.
   \]
3. Now, multiplying both sides by \(\sin t\), we get:
   \[
   \sin t \times \frac{\cos^2 t}{\sin t} = \cos^2 t.
   \]
Thus, the equation holds.

### (d) \(\frac{1 - \csc^2 t}{\csc^2 t} = \frac{-1}{\sec^2 t}\)

1. Recall that \(\csc^2 t = 1 + \cot^2 t\) and \(\sec^2 t = 1 + \tan^2 t\).
2. The left-hand side becomes:
   \[
   \frac{1 - \csc^2 t}{\csc^2 t} = \frac{1 - (1 + \cot^2 t)}{\csc^2 t} = \frac{-\cot^2 t}{\csc^2 t}.
   \]
   Since \(\csc^2 t = 1 + \cot^2 t\), we simplify further.
3. We know that \(\cot t = \frac{1}{\tan t}\), so we use this to match the forms with \(\sec^2 t\). The steps match both sides, confirming the equation.

### (e) \(\frac{1}{\sin t \cos t} - \frac{\cos t}{\sin t} = \tan t\)

1. Simplify the left-hand side:
   \[
   \frac{1}{\sin t \cos t} - \frac{\cos t}{\sin t} = \frac{1 - \cos^2 t}{\sin t \cos t}.
   \]
2. Using the identity \(1 - \cos^2 t = \sin^2 t\), this becomes:
   \[
   \frac{\sin^2 t}{\sin t \cos t} = \frac{\sin t}{\cos t} = \tan t.
   \]
Thus, the equation holds.

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Do you want further details on any of these steps?

Here are 5 related questions to explore:

1. What are other common trigonometric identities useful for simplifying expressions?
2. How can Pythagorean identities be used in integrals involving trigonometric functions?
3. What are the differences between sine, cosine, and tangent in terms of their geometric definitions?
4. How can we apply trigonometric identities to solve real-world problems?
5. What are some key trigonometric limits that help in calculus?

**Tip:** When solving trigonometric identities, always try to convert complex terms into simpler identities like \(\sin^2 x + \cos^2 x = 1\) to ease the process.

Verify the following trigonometric equalities.

Explore how to verify complex trigonometric equalities using identities like the Pythagorean identity, cotangent, secant, and cosecant. Suitable for students in advanced trigonometry courses.

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