If `A=tan^(-1) x ,x in R` then the value of sin 2A is

Options

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

Answer

To find the value of \( \sin 2A \) given that \( A = \tan^{-1} x \), we can follow these steps:

### Step 1: Understand the relationship
Given \( A = \tan^{-1} x \), we know that:
\[ \tan A = x \]

### Step 2: Use the double angle formula
We need to find \( \sin 2A \). We can use the double angle formula for sine:
\[ \sin 2A = 2 \sin A \cos A \]

### Step 3: Find \( \sin A \) and \( \cos A \)
To find \( \sin A \) and \( \cos A \), we can visualize \( A \) in a right triangle. In a right triangle:
- The opposite side (perpendicular) to angle \( A \) is \( x \).
- The adjacent side (base) to angle \( A \) is \( 1 \).

Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

Now we can find \( \sin A \) and \( \cos A \):
\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}} \]
\[ \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Substitute into the double angle formula
Now substitute \( \sin A \) and \( \cos A \) into the double angle formula:
\[ \sin 2A = 2 \sin A \cos A = 2 \left(\frac{x}{\sqrt{x^2 + 1}}\right) \left(\frac{1}{\sqrt{x^2 + 1}}\right) \]
\[ = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} \]

### Final Answer
Thus, the value of \( \sin 2A \) is:
\[ \sin 2A = \frac{2x}{x^2 + 1} \]

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