To solve this problem, let's rewrite $\sec(x)$ and $\csc(x)$ in terms of $\sin(x)$ and $\cos(x)$, and then simplify.

Step 1: Rewrite in terms of $\sin(x)$ and $\cos(x)$

1. The secant function $\sec(x)$ is defined as: $\sec(x) = \frac{1}{\cos(x)}$

2. The cosecant function $\csc(x)$ is defined as: $\csc(x) = \frac{1}{\sin(x)}$

So, the expression $\frac{\sec(x)}{\csc(x)}$ becomes: $\frac{\sec(x)}{\csc(x)} = \frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}}$

Step 2: Simplify the expression

To simplify, divide by $\frac{1}{\sin(x)}$ by multiplying with its reciprocal: $\frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{1} = \frac{\sin(x)}{\cos(x)}$

Step 3: Recognize the simplified form

The result $\frac{\sin(x)}{\cos(x)}$ is simply the tangent function: $\frac{\sin(x)}{\cos(x)} = \tan(x)$

Final Answer

$\frac{\sec(x)}{\csc(x)} = \tan(x)$

Would you like more details on any specific step or additional explanations?

Here are some related questions you might find useful:

1. How do you derive the basic trigonometric identities for $\sec(x)$ and $\csc(x)$?
2. What is the relationship between tangent and other trigonometric functions?
3. How do you simplify expressions involving $\sec(x)$ and $\csc(x)$ in other contexts?
4. Why does $\tan(x) = \frac{\sin(x)}{\cos(x)}$?
5. How would you simplify other combinations of trigonometric functions in terms of sine and cosine?

Tip: Memorizing the reciprocal and quotient identities of trigonometric functions can help simplify many problems more quickly.