https://file.aiquickdraw.com/mathbot/file/1728496303505yabukeb7.jpg

Let's walk through the solution to each part of the problem provided in the image.

### Given Function:
\[ f(x) = \frac{\tan(x) - 1}{\sec(x)} \]

#### (a) Using the Quotient Rule to Differentiate \( f(x) \):

You correctly applied the **Quotient Rule**, which is:
\[
f'(x) = \frac{(g(x) \cdot h'(x) - g'(x) \cdot h(x))}{h(x)^2}
\]
where \( g(x) = \tan(x) - 1 \) and \( h(x) = \sec(x) \).

After applying this, the derivative simplifies to:
\[
f'(x) = \frac{1 + \tan(x)}{\sec(x)}
\]

This part is correct.

#### (b) Simplify \( f(x) \) and Differentiate:

Now we are asked to rewrite the function in terms of \( \sin(x) \) and \( \cos(x) \).

1. Recall that:
   \[
   \tan(x) = \frac{\sin(x)}{\cos(x)} \quad \text{and} \quad \sec(x) = \frac{1}{\cos(x)}
   \]
   So, the function \( f(x) \) becomes:
   \[
   f(x) = \frac{\frac{\sin(x)}{\cos(x)} - 1}{\frac{1}{\cos(x)}}
   \]
2. Simplifying the numerator:
   \[
   \frac{\sin(x)}{\cos(x)} - 1 = \frac{\sin(x) - \cos(x)}{\cos(x)}
   \]
3. So, the function now is:
   \[
   f(x) = \frac{\frac{\sin(x) - \cos(x)}{\cos(x)}}{\frac{1}{\cos(x)}} = \sin(x) - \cos(x)
   \]
   Thus, we have simplified \( f(x) \) to:
   \[
   f(x) = \sin(x) - \cos(x)
   \]

Now we differentiate this simplified expression:
\[
f'(x) = \cos(x) + \sin(x)
\]

#### (c) Show the Equivalence of the Results:

We need to show that the derivative found in (a) is equivalent to the derivative found in (b).

1. From part (a):
   \[
   f'(x) = \frac{1 + \tan(x)}{\sec(x)}
   \]
   Rewriting this in terms of \( \sin(x) \) and \( \cos(x) \):
   \[
   f'(x) = \frac{1 + \frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}} = \frac{1 + \frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}} = \cos(x) + \sin(x)
   \]

This matches the derivative obtained in part (b). Hence, the answers are equivalent.

Would you like more details on any of these steps, or have any questions?

Here are 5 related questions for further practice:
1. What is the derivative of \( \sin(x) \cdot \cos(x) \)?
2. How would you apply the product rule to a function like \( \tan(x) \cdot \sec(x) \)?
3. Can you express \( \cot(x) \) and its derivative in terms of \( \sin(x) \) and \( \cos(x) \)?
4. How does the chain rule apply to the function \( \sec^2(x) \)?
5. How would you simplify \( \frac{1}{\sin(x) \cdot \sec(x)} \)?

**Tip**: When dealing with trigonometric functions, converting them to sine and cosine often simplifies the differentiation process.

Let the function f be defined by f(x) = (tan(x) - 1) / sec(x). Use the Quotient Rule to differentiate f(x). Simplify the expression for f(x) by writing it in terms of sin(x) and cos(x), then find f'(x). Show that your answers are equivalent.

Learn how to differentiate a trigonometric function defined by (tan(x) - 1) / sec(x) using the Quotient Rule and trigonometric identities. This problem involves rewriting the function in terms of sin(x) and cos(x), making it suitable for advanced high school calculus students.

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