`sin(tan^(-1)x),|x| le 1` is equal to :

Options

A. `(x)/(sqrt(1-x^(2)))`
B. `(1)/(sqrt(1-x^(2)))`
C. `(1)/(sqrt(1+x^(2)))`
D. `(x)/(sqrt(1+x^(2)))`

Explanation

To solve the problem \( \sin(\tan^{-1} x) \) where \( |x| \leq 1 \), we can follow these steps:

### Step 1: Let \( \theta = \tan^{-1} x \)
This means that \( \tan \theta = x \).

**Hint:** Recognize that \( \tan^{-1} x \) gives you an angle \( \theta \) whose tangent is \( x \).

### Step 2: Construct a right triangle
In a right triangle, if \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \), we can set:
- Opposite side = \( x \)
- Adjacent side = \( 1 \)

**Hint:** Visualize the triangle based on the definition of tangent.

### Step 3: Calculate the hypotenuse
Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

**Hint:** Remember that the hypotenuse is found by taking the square root of the sum of the squares of the other two sides.

### Step 4: Find \( \sin \theta \)
The sine of an angle in a right triangle is given by:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}} \]

**Hint:** Recall that sine is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

### Conclusion
Thus, we have:
\[ \sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2 + 1}} \]

**Final Answer:** The value of \( \sin(\tan^{-1} x) \) is \( \frac{x}{\sqrt{1 + x^2}} \).