`sin^(2)(2tan^(-1)sqrt((1+x)/(1-x)))="____","where "-1lexlt1`

Options

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

Answer

To solve the problem \( \sin^2(2 \tan^{-1}(\sqrt{\frac{1+x}{1-x}})) \) for \( -1 < x < 1 \), we will follow these steps:

### Step 1: Substitute \( x \) with \( \cos \theta \)
Let \( x = \cos \theta \). Then, we have:
\[ 1 + x = 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \]
\[ 1 - x = 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Thus, we can rewrite the expression under the square root:
\[ \sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{2 \cos^2\left(\frac{\theta}{2}\right)}{2 \sin^2\left(\frac{\theta}{2}\right)}} = \frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} = \cot\left(\frac{\theta}{2}\right) \]

### Step 2: Apply the tangent inverse function
Now substituting back, we have:
\[ \tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right) = \tan^{-1}\left(\cot\left(\frac{\theta}{2}\right)\right) = \frac{\pi}{2} - \frac{\theta}{2} \]

### Step 3: Calculate \( 2 \tan^{-1} \)
Now, we need to calculate:
\[ 2 \tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right) = 2\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \pi - \theta \]

### Step 4: Find \( \sin^2(\pi - \theta) \)
Using the sine function:
\[ \sin(\pi - \theta) = \sin(\theta) \]
Thus,
\[ \sin^2(2 \tan^{-1}(\sqrt{\frac{1+x}{1-x}})) = \sin^2(\theta) \]

### Step 5: Relate \( \sin^2(\theta) \) to \( x \)
Since \( x = \cos \theta \), we can use the Pythagorean identity:
\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - x^2 \]

### Final Result
Therefore, we conclude that:
\[ \sin^2(2 \tan^{-1}(\sqrt{\frac{1+x}{1-x}})) = 1 - x^2 \]