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

We use the logarithmic property $\ln(a^b) = b\ln(a)$, where $a = \cos(x)$ and $b = 2$. Thus,

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

So the expression becomes:

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

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

Recall the trigonometric identity $1 + \tan^2(x) = \sec^2(x)$. Using this identity, we have:

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

Now the expression becomes:

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

Step 3: Using the identity $\sec(x) = \frac{1}{\cos(x)}$

Next, we use the fact that $\sec(x) = \frac{1}{\cos(x)}$. Therefore, we can rewrite $\ln(\sec(x))$ as:

$\ln(\sec(x)) = \ln\left(\frac{1}{\cos(x)}\right) = -\ln(\cos(x)).$

Substituting this into the expression, we get:

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

This simplifies further to:

$e^x - 4\ln(\cos(x)).$

Final Simplified Expression:

Thus, the simplified expression is:

$e^x - 4\ln(\cos(x)).$

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Would you like a deeper explanation of any steps, or do you have further questions?

Related questions:

1. How does the identity $1 + \tan^2(x) = \sec^2(x)$ come from the Pythagorean identity?
2. Can the logarithmic properties used here be applied to other trigonometric functions?
3. What happens if we differentiate this simplified expression with respect to $x$?
4. How would this expression change if the original had $\sin^2(x)$ instead of $\cos^2(x)$?
5. Could we simplify the expression further if $x$ had specific values like $x = 0$?

Tip:

Always remember to apply the logarithmic properties carefully, especially when dealing with powers and products of functions inside the logarithm!