If `sin^(-1)((2a)/(1+a^(2))) + cos^(-1)((1-a^(2))/(1+a^(2)))=tan^(-1)((2x)/(1-x^(2)))`, where `a, x in]0,1[`, then the value of x is

Options

A. 0
B. `a/2`
C. a
D. `(2a)/(1-a^(2))`

Explanation

To solve the equation

\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right), \]

where \( a, x \in (0, 1) \), we can follow these steps:

### Step 1: Substitute \( a \) with \( \tan \theta \)

Let \( a = \tan \theta \). Then we have:

\[ \sin^{-1}\left(\frac{2\tan \theta}{1+\tan^2 \theta}\right) + \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \]

### Step 2: Use trigonometric identities

Using the identities:

\[ \sin(2\theta) = \frac{2\tan \theta}{1+\tan^2 \theta} \quad \text{and} \quad \cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}, \]

we can rewrite the equation as:

\[ \sin^{-1}(\sin(2\theta)) + \cos^{-1}(\cos(2\theta)). \]

### Step 3: Simplify the left-hand side

Since \( \sin^{-1}(\sin(2\theta)) + \cos^{-1}(\cos(2\theta)) = 2\theta \) (for \( 0 < 2\theta < \frac{\pi}{2} \)), we have:

\[ 2\theta = \tan^{-1}\left(\frac{2x}{1-x^2}\right). \]

### Step 4: Use the double angle formula for tangent

Using the double angle formula for tangent:

\[ \tan(2\theta) = \frac{2\tan \theta}{1-\tan^2 \theta}, \]

we can equate:

\[ \tan(2\theta) = \frac{2x}{1-x^2}. \]

### Step 5: Substitute \( \tan \theta \)

Since \( a = \tan \theta \), we can substitute \( \tan \theta \) back:

\[ \tan(2\theta) = \frac{2a}{1-a^2}. \]

### Step 6: Set the equations equal

Now we have:

\[ \frac{2a}{1-a^2} = \frac{2x}{1-x^2}. \]

### Step 7: Cross-multiply

Cross-multiplying gives:

\[ 2a(1-x^2) = 2x(1-a^2). \]

### Step 8: Simplify the equation

This simplifies to:

\[ 2a - 2ax^2 = 2x - 2ax^2. \]

### Step 9: Rearranging terms

Rearranging gives:

\[ 2a = 2x, \]

which leads to:

\[ x = a. \]

### Step 10: Substitute back to find \( x \)

Since \( a = \tan \theta \) and \( a \in (0, 1) \), we conclude that:

\[ x = \frac{2a}{1-a^2}. \]

Thus, the value of \( x \) is:

\[ \boxed{\frac{2a}{1-a^2}}. \]