Let's break down the expression \( \sqrt{1 - \cos^2(\tan^{-1}(x))} \) and simplify it.

### Step 1: Define \( \theta = \tan^{-1}(x) \)
Let \( \theta = \tan^{-1}(x) \), which means \( \tan(\theta) = x \). This defines an angle \( \theta \) such that its tangent is \( x \).

### Step 2: Use Trigonometric Identities
We know from trigonometric identities:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
Thus, \( 1 - \cos^2(\theta) = \sin^2(\theta) \).

### Step 3: Substitute and Simplify
Now substitute this into the original expression:
\[
\sqrt{1 - \cos^2(\theta)} = \sqrt{\sin^2(\theta)}
\]
This simplifies to:
\[
|\sin(\theta)|
\]
since the square root of a square is the absolute value.

### Step 4: Find \( \sin(\theta) \) in Terms of \( x \)
From our definition \( \theta = \tan^{-1}(x) \), we know the following trigonometric relationships for a right triangle where \( \tan(\theta) = x = \frac{\text{opposite}}{\text{adjacent}} \):
- Opposite side = \( x \)
- Adjacent side = 1
- Hypotenuse = \( \sqrt{x^2 + 1} \) (by the Pythagorean theorem)

Thus, \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}} \).

### Step 5: Final Answer
Therefore, \( |\sin(\theta)| = \left| \frac{x}{\sqrt{x^2 + 1}} \right| \).

So, the simplified expression is:
\[
\sqrt{1 - \cos^2(\tan^{-1}(x))} = \left| \frac{x}{\sqrt{x^2 + 1}} \right|
\]

Would you like more details on any of the steps?

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Here are 5 related questions:
1. How do you calculate trigonometric values for inverse functions?
2. What are some other key identities involving sine, cosine, and tangent?
3. Can this method be extended to other trigonometric expressions involving inverse functions?
4. How do you simplify expressions involving square roots and absolute values?
5. What are the practical applications of inverse trigonometric functions?

**Tip:** In trigonometric identities, using right triangle relationships helps simplify expressions involving inverse functions.

Learn how to simplify the expression √(1 - cos^2(tan^(-1)(x))) using trigonometric identities and the Pythagorean theorem. This solution involves understanding inverse trigonometric functions and the relationship between sine and cosine. Ideal for students in Grades 10-12.

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