To solve the expression:

\[
\frac{1}{\sec(x) - 1} + \frac{1}{\sec(x) + 1}
\]

We can follow these steps:

### Step 1: Combine the fractions
We first combine the two terms into a single fraction. To do that, we find a common denominator. The common denominator between \(\sec(x) - 1\) and \(\sec(x) + 1\) is:

\[
(\sec(x) - 1)(\sec(x) + 1)
\]

Now, using this common denominator, we can rewrite the expression as:

\[
\frac{1}{\sec(x) - 1} + \frac{1}{\sec(x) + 1} = \frac{(\sec(x) + 1) + (\sec(x) - 1)}{(\sec(x) - 1)(\sec(x) + 1)}
\]

### Step 2: Simplify the numerator
Simplify the numerator:

\[
(\sec(x) + 1) + (\sec(x) - 1) = \sec(x) + 1 + \sec(x) - 1 = 2\sec(x)
\]

### Step 3: Simplify the denominator
The denominator is a difference of squares:

\[
(\sec(x) - 1)(\sec(x) + 1) = \sec^2(x) - 1
\]

We can use the Pythagorean identity \(\sec^2(x) - 1 = \tan^2(x)\). Therefore, the denominator simplifies to \(\tan^2(x)\).

### Step 4: Final expression
Now we can rewrite the whole expression as:

\[
\frac{2\sec(x)}{\tan^2(x)}
\]

### Step 5: Further simplification
We know that \(\sec(x) = \frac{1}{\cos(x)}\) and \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). Thus:

\[
\frac{2\sec(x)}{\tan^2(x)} = \frac{2}{\cos(x)} \cdot \frac{\cos^2(x)}{\sin^2(x)} = \frac{2\cos(x)}{\sin^2(x)}
\]

This is the simplified form of the expression.

### Final Answer:

\[
\frac{2\cos(x)}{\sin^2(x)}
\]

Would you like further details or have any questions?

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Here are 5 related questions for further practice:
1. Simplify \( \frac{1}{\sec(x)-2} + \frac{1}{\sec(x)+2} \).
2. Prove the identity: \( \sec^2(x) - \tan^2(x) = 1 \).
3. Simplify \( \frac{1}{1 - \sin(x)} + \frac{1}{1 + \sin(x)} \).
4. Solve \( \frac{\sin(x)}{1 + \cos(x)} + \frac{\sin(x)}{1 - \cos(x)} \).
5. Simplify \( \frac{1}{\tan(x) + 1} + \frac{1}{\tan(x) - 1} \).

### Tip:
When simplifying trigonometric expressions, always check if any Pythagorean identities can help reduce the complexity.

Explore how to simplify the trigonometric expression 1/(secx-1) + 1/(secx+1) using key concepts like secant, tangent, and the Pythagorean identity. This detailed step-by-step solution is perfect for students in Grades 11-12 studying advanced trigonometry.

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