Prove that the given equation is not an identity by finding a value of x for which the two sides have different values.
cosecant xequals1 plus cotangent x

We are given the equation:

\[
\csc(x) = 1 + \cot(x)
\]

We want to find a value of \( x \) where the two sides of this equation are not equal, thus proving it is **not** an identity.

### Step 1: Recall the trigonometric definitions
- \( \csc(x) = \frac{1}{\sin(x)} \)
- \( \cot(x) = \frac{\cos(x)}{\sin(x)} \)

Thus, the equation becomes:
\[
\frac{1}{\sin(x)} = 1 + \frac{\cos(x)}{\sin(x)}
\]

### Step 2: Combine the right-hand side
To add \( 1 \) and \( \frac{\cos(x)}{\sin(x)} \), express \( 1 \) with a common denominator:
\[
1 = \frac{\sin(x)}{\sin(x)}
\]
Now, the right-hand side becomes:
\[
\frac{\sin(x)}{\sin(x)} + \frac{\cos(x)}{\sin(x)} = \frac{\sin(x) + \cos(x)}{\sin(x)}
\]

Thus, the equation simplifies to:
\[
\frac{1}{\sin(x)} = \frac{\sin(x) + \cos(x)}{\sin(x)}
\]

### Step 3: Cancel out \( \sin(x) \) on both sides
This leaves us with:
\[
1 = \sin(x) + \cos(x)
\]

### Step 4: Test for values of \( x \)
To prove this is **not** an identity, we need to find an \( x \) where \( 1 \neq \sin(x) + \cos(x) \).

Let’s test \( x = 0 \):
- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)

Thus, the equation becomes:
\[
1 \neq 0 + 1 = 1
\]
So at \( x = 0 \), both sides are equal. We need to try another value.

Let’s try \( x = \frac{\pi}{2} \):
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, the equation becomes:
\[
1 \neq 1 + 0 = 1
\]
Again, both sides are equal.

### Step 5: Find a value where the two sides differ
Now, let’s try \( x = \frac{\pi}{4} \) (45°):
- \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
- \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)

Now, substitute into the equation:
\[
1 \neq \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)
\]
\[
1 \neq \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}
\]
Since \( \sqrt{2} \approx 1.414 \), this proves that at \( x = \frac{\pi}{4} \), the two sides are not equal.

Thus, the given equation is **not** an identity.

### Conclusion:
The equation \( \csc(x) = 1 + \cot(x) \) is not an identity because it fails for values like \( x = \frac{\pi}{4} \), where the left side equals 1 and the right side equals \( \sqrt{2} \).

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. What is the definition of an identity in trigonometry?
2. How do you prove that a trigonometric equation is an identity?
3. What are other common trigonometric identities?
4. Can you explain the relationship between cosecant and cotangent in different quadrants?
5. How would this equation behave for values like \( x = \pi/6 \) or \( 3\pi/4 \)?

**Tip:** Always check multiple values to thoroughly test whether an equation is an identity.

Prove that the given equation is not an identity by finding a value of x for which the two sides have different values. cosecant x equals 1 plus cotangent x.

Explore the step-by-step solution to prove that the equation csc(x) = 1 + cot(x) is not a trigonometric identity by finding a value of x where the two sides differ. This problem covers trigonometric identities and equation testing, suitable for Grades 9-12.

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