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 will follow these steps:

### Step 1: Use the Identity for Inverse Trigonometric Functions
We know that:

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \]

for any \( y \) in the range of sine and cosine. Therefore, we can rewrite the left-hand side as:

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

### Step 2: Substitute the Values
From the identity, we can express the left-hand side as:

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

### Step 3: Use the Tangent Function
We know that:

\[ \tan\left(\frac{\pi}{2}\right) \text{ is undefined, but we can analyze the equation further.} \]

However, we can also use the relationship between the angles. Since \( \tan^{-1}(y) = \frac{\pi}{2} \) implies \( y \to \infty \), we need to find \( x \) such that:

\[ \frac{2x}{1-x^2} \to \infty \]

This occurs when the denominator approaches zero, i.e., \( 1 - x^2 = 0 \) or \( x = 1 \).

### Step 4: Relate the Two Sides
To find \( x \) in terms of \( a \), we can use the double angle formulas for tangent:

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

Thus, we can express:

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

### Step 5: Equate the Two Expressions
Setting the two expressions equal gives us:

\[ \tan^{-1}(x) = 2\tan^{-1}(a) \]

### Step 6: Solve for \( x \)
From the identity of tangent, we have:

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

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

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