https://file.aiquickdraw.com/mathbot/file/1726206634545jix0hk93.jpg

Let's work through this step-by-step to prove the trigonometric identity given in the image:

We are tasked with proving:

\[
\cot(x) + \tan(x) = \csc(x) \sec(x).
\]

### Step-by-step Proof:

1. Start by rewriting \(\cot(x)\) and \(\tan(x)\) in terms of sine and cosine:
   \[
   \cot(x) = \frac{\cos(x)}{\sin(x)}, \quad \tan(x) = \frac{\sin(x)}{\cos(x)}.
   \]
   Substituting these into the left-hand side of the equation:
   \[
   \cot(x) + \tan(x) = \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)}.
   \]

2. Now, find a common denominator for these two fractions:
   \[
   \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)} = \frac{\cos^2(x) + \sin^2(x)}{\sin(x) \cos(x)}.
   \]

3. Use the Pythagorean identity:
   \[
   \cos^2(x) + \sin^2(x) = 1.
   \]
   Therefore, the equation becomes:
   \[
   \frac{1}{\sin(x) \cos(x)}.
   \]

4. Notice that:
   \[
   \frac{1}{\sin(x)} = \csc(x), \quad \frac{1}{\cos(x)} = \sec(x).
   \]
   So, the expression simplifies to:
   \[
   \csc(x) \sec(x).
   \]

Thus, we have proven the identity:
\[
\cot(x) + \tan(x) = \csc(x) \sec(x).
\]

Would you like further details or have any questions about any part of this proof?

Here are five related questions:
1. What is the Pythagorean identity and why is it important in trigonometry?
2. How do you simplify trigonometric expressions with different denominators?
3. Can this identity be proven using different trigonometric identities?
4. What are the common strategies for proving trigonometric identities?
5. How do you derive \(\sec(x)\) and \(\csc(x)\) from basic trigonometric functions?

**Tip:** Always look for common identities like the Pythagorean identity to simplify complex trigonometric expressions quickly.

Prove the trigonometric identity cot(x) + tan(x) = csc(x) sec(x).

In this problem, we explore the trigonometric identity cot(x) + tan(x) = csc(x) sec(x). The solution walks through rewriting cotangent and tangent in terms of sine and cosine, applying the Pythagorean identity, and simplifying the expression.

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