The equation given is:

\[
\sec(x) = \tan(x) \cdot \csc(x)
\]

We can simplify or verify this trigonometric identity by expressing each trigonometric function in terms of sine and cosine.

1. **Rewrite the trigonometric functions:**

   - \(\sec(x) = \frac{1}{\cos(x)}\)
   - \(\tan(x) = \frac{\sin(x)}{\cos(x)}\)
   - \(\csc(x) = \frac{1}{\sin(x)}\)

2. **Substitute these into the equation:**

\[
\frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\sin(x)}
\]

3. **Simplify the right-hand side:**

\[
\frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\sin(x)} = \frac{\sin(x)}{\sin(x)\cos(x)} = \frac{1}{\cos(x)}
\]

4. **Conclusion:**

Both sides of the equation are equal, confirming that the identity is indeed correct.

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Do you want more details, or do you have any questions about this solution?

Here are 5 related questions you might find interesting:

1. How can \(\sec(x)\) and \(\csc(x)\) be expressed in terms of \(\tan(x)\) and \(\cot(x)\)?
2. What is the relationship between the basic trigonometric functions and their reciprocals?
3. How can you derive the Pythagorean identity from \(\sec^2(x)\) and \(\tan^2(x)\)?
4. Can you verify the identity \(\sec(x) - \tan(x) = \frac{\sin(x)}{1 + \cos(x)}\)?
5. What happens when you take the derivative of \(\sec(x)\) and \(\csc(x)\)?

**Tip:** When working with trigonometric identities, rewriting everything in terms of sine and cosine is a powerful strategy to simplify or prove identities.

Learn how to verify the trigonometric identity sec(x) = tan(x) * csc(x) with a step-by-step solution. This problem involves fundamental trigonometric concepts and identities. Suitable for students in Grades 11-12.

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