The given expression is:

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

Step-by-step simplification:

1. Use a trigonometric identity: Recall the identity for $\tan^2(x)$: $1 + \tan^2(x) = \sec^2(x)$ This allows us to replace $\ln(1 + \tan^2(x))$ with $\ln(\sec^2(x))$.

2. Simplify the logarithmic terms: The expression becomes: $e^x - \ln(\cos^2(x)) + \ln(\sec^2(x))$

   We also know that $\sec(x) = \frac{1}{\cos(x)}$, so $\sec^2(x) = \frac{1}{\cos^2(x)}$. Substituting this into the logarithmic term: $e^x - \ln(\cos^2(x)) + \ln\left(\frac{1}{\cos^2(x)}\right)$

3. Combine the logarithms: Using the property of logarithms $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$, the expression simplifies to: $e^x - \ln(\cos^2(x)) + \left( - \ln(\cos^2(x)) \right)$ This simplifies further as the logarithmic terms cancel out: $e^x$

Final Answer:

The simplified expression is: $e^x$

Would you like more details or have any questions?

Related questions:

1. Can you explain the logarithmic properties used in this simplification?
2. How does the trigonometric identity $1 + \tan^2(x) = \sec^2(x)$ work?
3. Can you show a different way to simplify this expression?
4. What are some other important logarithmic rules to remember?
5. How do exponential functions behave in relation to logarithms?

Tip:

Always remember to use trigonometric identities when simplifying expressions with logarithms and trigonometric functions!