We are tasked with simplifying the expression $\frac{\csc(-x)}{\sec(-x)}$ and verifying that it equals $-\cot(x)$.

Step 1: Use the properties of trigonometric functions

We know the following identities for negative angles:

• $\csc(-x) = -\csc(x)$ because $\csc(x) = \frac{1}{\sin(x)}$, and $\sin(-x) = -\sin(x)$.
• $\sec(-x) = \sec(x)$ because $\sec(x) = \frac{1}{\cos(x)}$, and $\cos(-x) = \cos(x)$.

Thus, we can rewrite the expression as: $\frac{\csc(-x)}{\sec(-x)} = \frac{-\csc(x)}{\sec(x)}$

Step 2: Express $\csc(x)$ and $\sec(x)$ in terms of $\sin(x)$ and $\cos(x)$

Using the definitions of $\csc(x)$ and $\sec(x)$, we have: $\frac{-\csc(x)}{\sec(x)} = \frac{-\frac{1}{\sin(x)}}{\frac{1}{\cos(x)}} = -\frac{\cos(x)}{\sin(x)}$

Step 3: Recognize the resulting expression

The expression $-\frac{\cos(x)}{\sin(x)}$ is the definition of $-\cot(x)$, because: $\cot(x) = \frac{\cos(x)}{\sin(x)}$ Therefore: $-\frac{\cos(x)}{\sin(x)} = -\cot(x)$

Final Result:

$\frac{\csc(-x)}{\sec(-x)} = -\cot(x)$ This confirms the identity is true.

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Would you like further clarification or have any additional questions on this topic?

Related Questions:

1. What is the proof of the identity $\csc(x) = \frac{1}{\sin(x)}$?
2. How does the property $\cos(-x) = \cos(x)$ relate to even functions in trigonometry?
3. What is the relationship between $\cot(x)$ and other trigonometric functions like $\tan(x)$?
4. How would you simplify $\frac{\sec(x)}{\csc(x)}$?
5. What are the properties of even and odd functions in trigonometry?

Tip: When working with trigonometric identities involving negative angles, remember the key rules that $\sin(-x) = -\sin(x)$ (odd function) and $\cos(-x) = \cos(x)$ (even function). These can help simplify most expressions.