Consider the expression $ e^x-ln(cos^2(x))+ln(1+tan^2(x))$ and then simplify it into a single term

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.

Consider the expression $ e^x - \ln(\cos^2(x)) + \ln(1 + \tan^2(x))$ and then simplify it into a single term.

Learn how to simplify the expression e^x - ln(cos^2(x)) + ln(1 + tan^2(x)) step by step. This problem involves exponential functions, logarithmic identities, and trigonometric identities like the Pythagorean identity. Suitable for advanced high school students in Grades 11-12.

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