We are given the expression:

$e^x - \ln(\cos^2(x)) + \ln(1 + \tan^2(x))$

Let's break it down and simplify it step by step.

Step 1: Simplifying $\ln(1 + \tan^2(x))$

Recall that $1 + \tan^2(x) = \sec^2(x)$ from the Pythagorean identity. So we have:

$\ln(1 + \tan^2(x)) = \ln(\sec^2(x))$

Since $\ln(a^b) = b\ln(a)$, this becomes:

$\ln(\sec^2(x)) = 2\ln(\sec(x))$

Step 2: Simplifying $- \ln(\cos^2(x))$

Using the logarithmic property $\ln(a^b) = b\ln(a)$, we can write:

$- \ln(\cos^2(x)) = -2\ln(\cos(x))$

Step 3: Combine the logarithmic terms

Now, let's combine the two logarithmic terms:

$- 2\ln(\cos(x)) + 2\ln(\sec(x))$

Since $\sec(x) = \frac{1}{\cos(x)}$, we know that $\ln(\sec(x)) = -\ln(\cos(x))$. Thus:

$2\ln(\sec(x)) = -2\ln(\cos(x))$

So the entire logarithmic part becomes:

$- 2\ln(\cos(x)) + (-2\ln(\cos(x))) = 0$

Step 4: Final expression

After simplifying the logarithmic terms to 0, the remaining expression is just:

$e^x$

Final Simplified Expression:

$e^x$

This is the fully simplified version of the given expression.

---

Would you like more details or explanations about this simplification?

Here are five related questions you might find useful:

1. How can logarithmic properties be used to simplify more complex expressions?
2. What are the Pythagorean identities, and how are they used in trigonometry?
3. How does the logarithmic identity $\ln(a^b) = b\ln(a)$ apply to general logarithmic functions?
4. What are some common simplifications involving $\tan(x)$ and $\sec(x)$?
5. How can exponential functions like $e^x$ be manipulated in calculus?

Tip: Always check for trigonometric identities like the Pythagorean identity, as they often simplify complex trigonometric expressions easily.