If `tan^(-1). (sqrt((1+x^(2))) - sqrt((1-x^(2))))/(sqrt((1+x^(2)))+sqrt((1-x^(2))))=alpha" , then " x^(2) ` is

Options

A. `cos 2 alpha`
B. `sin 2 alpha`
C. `tan 2 alpha `
D. `cot 2 alpha `

Answer

To solve the given problem, we will follow these steps:

### Step 1: Set up the equation
We start with the equation given in the problem:
\[ \tan^{-1}\left(\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha \]

### Step 2: Rewrite in terms of tangent
Using the property of the inverse tangent function, we can rewrite the equation as:
\[ \tan(\alpha) = \frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \]

### Step 3: Simplify the right-hand side
To simplify the right-hand side, we can multiply the numerator and denominator by \(\sqrt{1+x^2} - \sqrt{1-x^2}\):
\[ \tan(\alpha) = \frac{(\sqrt{1+x^2} - \sqrt{1-x^2})^2}{(1+x^2) - (1-x^2)} \]

### Step 4: Expand the numerator
Expanding the numerator:
\[ (\sqrt{1+x^2} - \sqrt{1-x^2})^2 = (1+x^2) + (1-x^2) - 2\sqrt{(1+x^2)(1-x^2)} = 2 - 2\sqrt{(1+x^2)(1-x^2)} \]
The denominator simplifies to:
\[ (1+x^2) - (1-x^2) = 2x^2 \]

### Step 5: Combine the results
Thus, we have:
\[ \tan(\alpha) = \frac{2 - 2\sqrt{(1+x^2)(1-x^2)}}{2x^2} = \frac{1 - \sqrt{(1+x^2)(1-x^2)}}{x^2} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ x^2 \tan(\alpha) = 1 - \sqrt{(1+x^2)(1-x^2)} \]

### Step 7: Square both sides
Squaring both sides:
\[ (x^2 \tan(\alpha))^2 = (1 - \sqrt{(1+x^2)(1-x^2)})^2 \]

### Step 8: Expand both sides
Expanding the right-hand side:
\[ (x^2 \tan(\alpha))^2 = 1 - 2\sqrt{(1+x^2)(1-x^2)} + (1+x^2)(1-x^2) \]
The right-hand side simplifies to:
\[ 1 - 2\sqrt{(1+x^2)(1-x^2)} + 1 - x^4 \]

### Step 9: Rearranging terms
Rearranging gives:
\[ x^4 \tan^2(\alpha) + x^4 = 2 - 2\sqrt{(1+x^2)(1-x^2)} \]

### Step 10: Solve for \(x^2\)
From here, we can isolate \(x^2\):
\[ x^4(\tan^2(\alpha) + 1) = 2 - 2\sqrt{(1+x^2)(1-x^2)} \]
Using the identity \(\tan^2(\alpha) + 1 = \sec^2(\alpha)\), we can express \(x^2\) in terms of \(\alpha\):
\[ x^2 = \frac{2\sin(2\alpha)}{\sec^2(\alpha)} \]

### Final Result
Thus, we find:
\[ x^2 = \sin(2\alpha) \]