We are given the expression:

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

Let's simplify this step by step.

Step 1: Use the identity for $\tan^2(x)$

Recall the Pythagorean identity for tangent: $1 + \tan^2(x) = \sec^2(x)$ So, we can replace $\ln(1 + \tan^2(x))$ with $\ln(\sec^2(x))$. The expression now becomes:

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

Step 2: Simplify the logarithmic terms

We know that $\sec(x) = \frac{1}{\cos(x)}$, so:

$\sec^2(x) = \frac{1}{\cos^2(x)}$

Thus, $\ln(\sec^2(x)) = \ln\left(\frac{1}{\cos^2(x)}\right)$. Using logarithmic properties:

$\ln\left(\frac{1}{\cos^2(x)}\right) = -\ln(\cos^2(x))$

So the expression now becomes:

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

Step 3: Combine the terms

Notice that the two logarithmic terms cancel each other out:

$e^x$

Final result:

The simplified expression is:

$e^x$

Would you like further explanation or have any questions about the simplification? Here are some related questions:

1. How do logarithmic identities simplify expressions like this one?
2. What are the properties of exponential and logarithmic functions that apply here?
3. How does the identity $\sec^2(x) = 1 + \tan^2(x)$ arise from basic trigonometry?
4. What are other trigonometric identities involving $\sin(x)$ and $\cos(x)$?
5. How does simplifying logarithmic expressions help in calculus or differential equations?

Tip: Always use logarithmic identities like $\ln(a/b) = \ln(a) - \ln(b)$ to break down expressions effectively.