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.