To determine the Maclaurin series for $\tan(\ln(1 + x))$ up to the $x^3$ term in part (c), we can proceed as follows:

1. Use Previous Results: In parts (a) and (b), we obtained the Maclaurin series expansions for $\cos(\ln(1 + x))$ and $\sin(\ln(1 + x))$. For this question, we need the Maclaurin series for $\tan(\ln(1 + x)) = \frac{\sin(\ln(1 + x))}{\cos(\ln(1 + x))}$.

2. Maclaurin Series for $\sin(\ln(1 + x))$ and $\cos(\ln(1 + x))$: From part (a), we have: $\cos(\ln(1 + x)) = 1 - \frac{1}{2}x^2 + \frac{1}{2}x^3 - \frac{5}{12}x^4 + \cdots$ From part (b), we have: $\sin(\ln(1 + x)) = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \cdots$

3. Form the Quotient $\frac{\sin(\ln(1 + x))}{\cos(\ln(1 + x))}$: We want to expand $\tan(\ln(1 + x))$ using the quotient: $\tan(\ln(1 + x)) = \frac{\sin(\ln(1 + x))}{\cos(\ln(1 + x))}.$

4. Use Series Division: Perform the division of $\sin(\ln(1 + x))$ by $\cos(\ln(1 + x))$, keeping terms up to $x^3$:

   $\tan(\ln(1 + x)) = \frac{x - \frac{1}{2}x^2 + \frac{1}{6}x^3}{1 - \frac{1}{2}x^2 + \frac{1}{2}x^3}.$

   To divide, write this as: $\tan(\ln(1 + x)) = \left(x - \frac{1}{2}x^2 + \frac{1}{6}x^3\right) \cdot \left(1 + \frac{1}{2}x^2 - \frac{1}{2}x^3\right).$

5. Multiply Out Terms up to $x^3$: Expanding the product: $\tan(\ln(1 + x)) = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{2}x^3 - \frac{1}{4}x^4 + \cdots$ Simplify terms up to $x^3$: $\tan(\ln(1 + x)) = x - \frac{1}{2}x^2 + \frac{2}{3}x^3 + \cdots$

Thus, the Maclaurin series for $\tan(\ln(1 + x))$ up to the $x^3$ term is: $\tan(\ln(1 + x)) = x - \frac{1}{2}x^2 + \frac{2}{3}x^3 + \cdots$

Would you like further details or have any questions?

Here are some related questions to explore:

1. How is the Maclaurin series for $\ln(1 + x)$ derived?
2. What is the general approach to finding Maclaurin series for composite functions?
3. How does term-by-term differentiation affect a Maclaurin series?
4. Can the series expansion be applied to functions other than trigonometric and logarithmic functions?
5. How do higher-order terms affect the accuracy of approximations in Maclaurin series?

Tip: In series division, remember to multiply each term in the numerator by each term in the denominator (up to the desired degree), simplifying the result term by term.