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.